Let $m > n$. By the Pigeonhole Principle, there can be no injection from $S$ to $T$ when $\left|{S}\right| > \left|{T}\right|$. Once $\left|{T}\right|$ elements of $S$ have been used up, there is no element of $T$ left for the remaining elements of $S$ to be mapped to such that they all still map to different elements of $T$. Let $m = 0$. The only injection from $\varnothing \to T$ is $\varnothing \times T$ which is $\varnothing$. So if $m = 0$ there is $1 = n! / n!$ injection. Let $0 < m \le n$. As in the proof of Cardinality of Set of All Mappings, we can assume that $S = \N_m$ and $T = \N_n$. For each $k \in \left[{1 \,.\,.\, n}\right]$, let $\mathbb H \left({k, n}\right)$ be the set of all injections from $\N_k$ to $\N_n$. The proof now proceeds by induction. Let: :$\displaystyle \mathbb S = \left\{{k \in \left[{1 \,.\,.\, n}\right]: \left|{\mathbb H \left({k, n}\right)}\right| = \frac {n!} {\left({n - k}\right)!}}\right\}$ === Basis for the Induction === Let $k = 1$. From Cardinality of Set of All Mappings, there are $n^1 = n$ different mappings from $S$ to $T$. From Mapping from Singleton is Injection, each one of these $n$ mappings is an injection. Thus: :$\displaystyle \left|{\mathbb H \left({1, n}\right)}\right| = n = \frac {n!} {\left({n - 1}\right)!}$ and so it follows that: :$1 \in \mathbb S$. This is the basis for the induction. === Induction Hypothesis === We suppose that: :$\displaystyle \left|{\mathbb H \left({k, n}\right)}\right| = \frac {n!} {\left({n - k}\right)!}$. This is the induction hypothesis. We need to show that: :$\displaystyle \left|{\mathbb H \left({k+1, n}\right)}\right| = \frac {n!} {\left({n - \left({k+1}\right)}\right)!}$. === Induction Step === This is the induction step: Let $k \in \mathbb S$ such that $k < n$. Let $\rho: \mathbb H \left({k + 1, n}\right) \to \mathbb H \left({k, n}\right)$ be the mapping defined by: :$\forall f \in \mathbb H \left({k + 1, n}\right): \rho \left({f}\right) =$ the restriction of $f$ to $\N_k$ Given that $g \in \mathbb H \left({k, n}\right)$ and $a \in \N_n - g \left({\N_k}\right)$, let $g_a: \N_{k + 1} \to \N_n$ be the mapping defined as: :$g_a \left({x}\right) = \begin{cases} g \left({x}\right) & : x \in \N_k \\ a & : x = k \end{cases}$ Now $g$ is an injection as $g \in \mathbb H \left({k, n}\right)$, and as $g_a \left({a}\right) \notin g \left({\N_k}\right)$ it follows that $g_a$ is also an injection. Hence: :$g_a \in \mathbb H \left({k + 1, n}\right)$ It follows from the definition of $\rho$ that: :$\rho^{-1} \left({\left\{{g}\right\}}\right) = \left\{{g_a: a \in \N_n - g \left({\N_k}\right)}\right\}$ Since $g$ is an injection, $g \left({\N_k}\right)$ has $k$ elements. Therefore $\N_n - g \left({\N_k}\right)$ has $n - k$ elements by Cardinality of Complement. As $G: a \to g_a$ is clearly a bijection from $\N_n - g \left({\N_k}\right)$ onto $\rho^{-1} \left({\left\{{g}\right\}}\right)$, that set has $n - k$ elements. Clearly: :$\left\{{\rho^{-1} \left({\left\{{g}\right\}}\right): g \in \mathbb H \left({k, n}\right)}\right\}$ is a partition of $\mathbb H \left({k + 1, n}\right)$. Therefore by Number of Elements in Partition: :$\left|{\mathbb H \left({k + 1, n}\right)}\right| = \left({n - k}\right) \dfrac {n!} {\left({n - k}\right)!} = \dfrac {n!} {\left({\left({n - k}\right) - 1}\right)!}$ as $k \in \mathbb S$. But: :$\left({n - k}\right) - 1 = n - \left({k + 1}\right)$ So: :$k + 1 \in \mathbb S$ By induction: :$\mathbb S = \left[{1 \,.\,.\, n}\right]$ and in particular: :$m \in \mathbb S$ {{qed}}