Let $\alpha$ be an algebraic number such that $\alpha \ne 0$ and $\alpha \ne 1$. Let $\beta$ be an algebraic number such that some value of $\alpha^\beta$ is algebraic. The result will follow if we can show that $\beta \in \Q$. We treat only the special case that $\alpha, \beta \in \R$ and $\alpha > 0$, assuming that $\map \exp {\beta \ln \alpha}$ is algebraic for the real value of $\ln \alpha$. Observe that $\alpha^{s_1 + s_2 \beta}$ is an algebraic number for all integers $s_1$ and $s_2$. To establish the result, it is enough to show that there are two distinct pairs of integers $\tuple {s_1, s_2}$ and $\tuple {s'_1, s'_2}$ for which: :$s_1 + s_2 \beta = s'_1 + s'_2 \beta$  since that implies: :$\beta = \dfrac{s_1 - s'_1} {s'_2 - s_2} \in \Q$ We will choose $S$ sufficiently large and show such pairs exist with $0 \le s_1, s_2, s'_1, s'_2 < S$. === Lemma 1 === {{:Gelfond-Schneider Theorem/Lemma 1}} === Lemma 2 === {{:Gelfond-Schneider Theorem/Lemma 2}} === Lemma 3 === {{:Gelfond-Schneider Theorem/Lemma 3}} === Lemma 4 === {{:Gelfond-Schneider Theorem/Lemma 4}} === Construction of the matrix $\boldsymbol M$ and outline of the proof === Let $c$ be a sufficiently large real number (which will be specified in due course). Consider integers $L_0, L_1, S$ each $\ge 2$. Let $L = \paren {L_0 + 1} \paren {L_1 + 1}$. Observe that we can find such $L_0, L_1, S$ (and we do so) with: :$c L_0 \ln S \le L$, $c L_1 S \le L$, $L \le S^2$ For example, take $S$ large, and: :$L_0 = \floor {S \ln S}, L_1 = \floor {\dfrac S {2 \ln S} }$ Note that we could take $c = \ln \ln S$. Consider the matrix $M$ described as follows. Consider some arrangement $\tuple {\map {s_1} i, \map {s_2} i}$ of the $S^2$ integer pairs $\tuple {s_1, s_2}$ with $0 \le s_1 < S$ and $0 \le s_2 < S$. Also, consider some arrangement $\tuple {\map u j, \map v j}$, with $1 \le j \le L$, of the integer pairs $\tuple {u, v}$ where $0 \le u \le L_0$ and $0 \le v \le L_1$. Then we define: :$M = \sqbrk {\paren {\map {s_1} i + \map {s_2} i \beta}^{\map u j} \paren {\alpha^{\map {s_1} i + \map {s_2} i \beta} }^{\map v j} }$ so that $M$ is a $S^2 \times L$ matrix. The idea of the proof is to: :$(1): \quad$ Consider the determinant $\Delta$ of an arbitrary $L \times L$ submatrix of $M$ (any one will do). :$(2): \quad$ Use Lemma 3 to obtain an upper bound $B_1$ for the absolute value of $\Delta$ (or, more specifically, an upper bound for $\ln \size \Delta$). :$(3): \quad$ Use Lemma 4 to motivate that if $\Delta \ne 0$, then $\size \Delta \ge B_2$ for some $B_2 > B_1$. :$(4): \quad$ Conclude that $\Delta$ must be $0$ and, hence, the rank of $M$ is less than $L$. :$(5): \quad$ Take a linear combination of the columns of $M$ to obtain an $\map F t$ as in Lemma 1 with less than $L$ roots but with $\map F {\map {s_1} i + \map {s_2} i \beta} = 0$ for $1 \le i \le L$. :$(6): \quad$ Conclude that $\beta$ is rational as required. === Upper bound for $\size \Delta$ === We have not specified the arrangements defining $\tuple {\map {s_1} i, \map {s_2} i}$ and $\tuple {\map u j, \map v j}$. Therefore it suffices to consider $\Delta = \det \sqbrk {\map {f_j} {\zeta_i} }$ where: :$\map {f_j} z = z^{\map u j} \alpha^{\map v j z} \quad \paren {1 \le j \le L}$ and: :$\zeta_i = \map {s_1} i + \map {s_2} i \beta \quad \paren {1 \le i \le L}$ Observe that $\map u j$ is a non-negative integer for each $j$. Also, $\alpha^{\map v j z} = \map \exp {\map v j z \ln \alpha}$, where $\ln \alpha$ denotes the real value of the logarithm. Hence, $\map {f_j} z$ is uniquely defined. Then $\map {f_j} z$ represents an entire function for each $j$. Observe that: :$\size {e^{z_1 z_2} } = e^{\map \Re {z_1 z_2} } \le e^{\size {z_1 z_2} } = e^{\size {z_1} \size {z_2} }$ for all complex numbers $z_1$ and $z_2$. {{explain|Prove the above, or link to such a proof}} Hence, for any $R > 0$: :$\size {f_j}_R \le R^{\map u j} e^{\map v j R \size {\ln \alpha} }$ We use Lemma 3 with $r = S \paren {1 + \size {\beta} }$ and $R = e^2 r$. Then for some constant $c_1 > 0$, we obtain: {{begin-eqn}} {{eqn | l = \ln \size \Delta       | o = \le       | r = -L \paren {L - 1} + \ln L! + L \max_{1 \mathop \le j \mathop \le L} \set {\ln \size {f_j}_R}       | c =  }} {{eqn | o = \le       | r = -L \paren {L - 1} + L \ln L + L L_0 \ln R + L L_1 R \size {\ln \alpha}       | c =  }} {{eqn | o = \le       | r = -L^2 + c_1 \paren {L L_0 \ln S + L L_1 S}       | c =  }} {{end-eqn}} The constant $c_1$ above is independent of $c$. Therefore, if $c$ is sufficiently large (that is, $c \ge 4 c_1$), then: :$\ln \size \Delta \le -\dfrac {L^2} 2$ === Lower bound for $\size \Delta$ if $\Delta \ne 0$ === Suppose now that $T'$ is a positive rational integer for which $T' \alpha, T' \beta$ and $T'\alpha^\beta$ are all algebraic integers. Then $T = \paren {T'}^{L_0 + 2 S L_1}$ has the property that $T$ times any element of $M$ (and, hence, $T$ times any element of the matrix defining $\Delta$) is an algebraic integer. Therefore, by Lemma 4, if $\Delta \ne 0$, then there is a conjugate of $\Delta$ with absolute value $\ge T^{-L} = \paren {T'}^{-L L_0 - 2 S L L_1}$. It may be reasonable to expect that a similar inequality might hold for $\size \Delta$ itself (rather than for the absolute value of a conjugate of $\Delta$). It will be shown later that if $\Delta \ne 0$, then there is indeed a constant $c_2$ (independent of $c$) for which: :$(A) \quad \ln \size \Delta \ge -c_2 \paren {L L_0 \ln S + S L L_1}$ === Conclusion === We see that, for $c$ sufficiently large ($c \ge 8 c_2$ will do), our upper bound for $\ln \size \Delta$ would contradict $(A)$, hence we obtain that $\Delta = 0$. Since $\Delta = \det \sqbrk {\map {f_j} {\zeta_i} }$ as defined above, we get that the columns of $\sqbrk {\map {f_j} {\zeta_i} }$ must be linearly dependent over the real numbers. In other words, there exist real numbers $b_j$, not all $0$, such that: :$\displaystyle \sum_{j \mathop = 1}^L b_j \map {f_j} {\zeta_i} = 0$ for $1 \le i \le L$ We now order the pairs $\tuple {u, v}$ in such a way that $\tuple {u, v} \le \tuple {u', v'}$ {{iff}} $v < v'$, or $v = v'$ and $u \le u'$. We deduce that: :$\displaystyle \sum_{v \mathop = 0}^{L_1} \sum_{u \mathop = 0}^{L_0} b_{\paren {L_0 + 1} v + u + 1} \zeta_i^u \alpha^{v \zeta_i} = 0$ for $1 \le i \le L$ But: :$\displaystyle \sum_{v \mathop = 0}^{L_1} \sum_{u \mathop = 0}^{L_0} b_{\paren {L_0 + 1} v + u + 1} \zeta_i^u \alpha^{v \zeta_i} = \sum_{v \mathop = 0}^{L_1} \map {a_v} t e^{w_v t}$ where: :$\displaystyle \map {a_v} t = \sum_{u \mathop = 0}^{L_0} b_{\paren {L_0 + 1} v + u + 1} t^u, w_v = v \ln \alpha$ and: :$t = \zeta_i = \map {s_1} i + \map {s_2} i \beta$ Each of the $L$ values of $\zeta_i$ is a root of the function $\displaystyle \sum_{v \mathop = 0}^{L_1} \map {a_v} t e^{w_v t}$. Since some $b_j \ne 0$, we deduce from Lemma 1 that there are at most $L_0 \paren {L_1 + 1} + \paren {L_1 + 1} - 1 < L$ distinct real roots. Therefore, two roots $\zeta_i$ must be the same, and we can conclude that: :$\map {s_1} i + \map {s_2} i \beta = \map {s_1} {i'} + \map {s_2} {i'} \beta$ for some $i, i'$ with $1 \le i < i' \le L$. On the other hand, the pairs $\tuple {\map {s_1} i, \map {s_2} i}$ and $\tuple {\map {s_1} {i'}, \map {s_2} {i'} }$ are necessarily distinct. So we can conclude that $\beta$ is rational, completing the proof. === Proof of inequality $(A)$ === To finish the proof, all we need to do is to show that if $\Delta \ne 0$, then $(A)$ holds. Assume $\Delta \ne 0$ and recall that $T^L \Delta$ is an algebraic integer, where $T=(T')^{L_0+2SL_1}$ and $T'$ is a positive integer (depending only on $\alpha$, $\beta$, and $\alpha^\beta$). For an algebraic number $w$, we denote by $\norm w$ the maximum of the absolute value of a conjugate of $w$. Observe that: :$\norm {w + w'} \le \norm w + \norm {w'}$, $\norm {w w'} \le \norm w \norm {w'}$ and: :$\norm {\lambda w} = \size \lambda \norm w$ whenever $w, w'$ are algebraic numbers and $\lambda \in \Q$. Then we obtain that: :$\norm {T^L \Delta} = T^L \norm \Delta \le T^L L! S^{L_0 L} \paren {1 + \norm \beta}^{L_0 L} \paren {1 + \norm \alpha}^{S L_1 L} \paren {1 + \norm {\alpha^\beta} }^{S L_1 L}$ We have that $T^L \Delta$ is an algebraic integer in $\map Q {\alpha, \beta, \alpha^\beta}$. So we can deduce that $T^L \Delta$ is a root of an irreducible monic polynomial $\map g x$ of degree $N$ where $N$ is the product of the degrees of the minimal polynomials for $\alpha, \beta, \alpha^\beta$. Note that each root of $\map g x$ is a conjugate of $T^L \Delta$. Since: :the product of all the roots of $\map g x$ has absolute value $\size {\map g 0} \ge 1$ and: :each root of $\map g x$ has absolute value $\le \norm {T^L \Delta}$ it follows that: {{begin-eqn}} {{eqn | l = \size {T^L \Delta}       | o = \ge       | r = \dfrac {\size {\map g 0} } {\norm {T^L \Delta}^{N - 1} }       | c =  }} {{eqn | o = \ge       | r = T^{-\paren {N - 1} L} \paren {L!}^{-N} S^{-N L_0 L} \paren {1 + \norm \beta}^{-N L_0 L} \paren {1 + \norm \alpha}^{-N S L_1 L} \paren {1 + \norm {\alpha^\beta} }^{-N S L_1 L}       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \ln \size \Delta       | o = \ge       | r = -N L \ln T - N L \ln L - N L_0 L \ln S - N L_0 L \map \ln {1 + \norm \beta}       | c =  }} {{eqn | o =        | r = - N S L_1 L \map \ln {1 + \norm \alpha} - N S L_1 L \map \ln {1 + \norm {\alpha^\beta} }       | c =  }} {{end-eqn}} Recall that $T = \paren {T'}^{L_0 + 2 S L_1}$ and that $T'$ and $N$ are constants depending only on $\alpha$ and $\beta$. It follows that $(A)$ holds, and the proof for $\alpha > 0$ and $\beta$ real follows. {{qed}} {{Namedfor|Alexander Osipovich Gelfond|name2 = Theodor Schneider|cat = Gelfond|cat2 = Schneider}}