Let $x_1 , \ldots, x_n$ be elements of $\mathfrak m$ whose images in $\mathfrak m / \mathfrak m^2$ form a basis. Consider the {\it Koszul complex} on $x_1, \ldots, x_n$. This is the complex $$ 0 \to \wedge^n R^n \to \wedge^{n-1} R^n \to \wedge^{n-2} R^n \to \ldots \to \wedge^i R^n \to \ldots \to R^n \to R $$ with maps given by $$ e_{j_1} \wedge \ldots \wedge e_{j_i} \longmapsto \sum_{a = 1}^i (-1)^{i + 1} x_{j_a} e_{j_1} \wedge \ldots \wedge \hat e_{j_a} \wedge \ldots \wedge e_{j_i} $$ It is easy to see that this is a complex $K_{\bullet}(R, x_{\bullet})$. Note that the cokernel of the last map of $K_{\bullet}(R, x_{\bullet})$ is $\kappa$ by Lemma \ref{lemma-NAK} part (8). \medskip\noindent If $\kappa$ has finite projective dimension $d$, then we can find a resolution $F_{\bullet} \to \kappa$ by finite free $R$-modules of length $d$ (Lemma \ref{lemma-what-kind-of-resolutions-Noetherian-local}). By Lemma \ref{lemma-add-trivial-complex} we may assume all the maps in the complex $F_{\bullet}$ have the property that $\Im(F_i \to F_{i-1}) \subset \mathfrak m F_{i-1}$, because removing a trivial summand from the resolution can at worst shorten the resolution. By Lemma \ref{lemma-compare-resolutions} we can find a map of complexes $\alpha : K_{\bullet}(R, x_{\bullet}) \to F_{\bullet}$ inducing the identity on $\kappa$. We will prove by induction that the maps $\alpha_i : \wedge^i R^n = K_i(R, x_{\bullet}) \to F_i$ have the property that $\alpha_i \otimes \kappa : \wedge^i \kappa^n \to F_i \otimes \kappa$ are injective. This shows that $F_n \not = 0$ and hence $d \geq n$ as desired. \medskip\noindent The result is clear for $i = 0$ because the composition $R \xrightarrow{\alpha_0} F_0 \to \kappa$ is nonzero. Note that $F_0$ must have rank $1$ since otherwise the map $F_1 \to F_0$ whose cokernel is a single copy of $\kappa$ cannot have image contained in $\mathfrak m F_0$. \medskip\noindent Next we check the case $i = 1$ as we feel that it is instructive; the reader can skip this as the induction step will deduce the $i = 1$ case from the case $i = 0$. We saw above that $F_0 = R$ and $F_1 \to F_0 = R$ has image $\mathfrak m$. We have a commutative diagram $$ \begin{matrix} R^n & = & K_1(R, x_{\bullet}) & \to & K_0(R, x_{\bullet}) & = & R \\ & & \downarrow & & \downarrow & & \downarrow \\ & & F_1 & \to & F_0 & = & R \end{matrix} $$ where the rightmost vertical arrow is given by multiplication by a unit. Hence we see that the image of the composition $R^n \to F_1 \to F_0 = R$ is also equal to $\mathfrak m$. Thus the map $R^n \otimes \kappa \to F_1 \otimes \kappa$ has to be injective since $\dim_\kappa (\mathfrak m / \mathfrak m^2) = n$. \medskip\noindent Let $i \geq 1$ and assume injectivity of $\alpha_j \otimes \kappa$ has been proved for all $j \leq i - 1$. Consider the commutative diagram $$ \begin{matrix} \wedge^i R^n & = & K_i(R, x_{\bullet}) & \to & K_{i-1}(R, x_{\bullet}) & = & \wedge^{i-1} R^n \\ & & \downarrow & & \downarrow & & \\ & & F_i & \to & F_{i-1} & & \end{matrix} $$ We know that $\wedge^{i-1} \kappa^n \to F_{i-1} \otimes \kappa$ is injective. This proves that $\wedge^{i-1} \kappa^n \otimes_{\kappa} \mathfrak m/\mathfrak m^2 \to F_{i-1} \otimes \mathfrak m/\mathfrak m^2$ is injective. Also, by our choice of the complex, $F_i$ maps into $\mathfrak mF_{i-1}$, and similarly for the Koszul complex. Hence we get a commutative diagram $$ \begin{matrix} \wedge^i \kappa^n & \to & \wedge^{i-1} \kappa^n \otimes \mathfrak m/\mathfrak m^2 \\ \downarrow & & \downarrow \\ F_i \otimes \kappa & \to & F_{i-1} \otimes \mathfrak m/\mathfrak m^2 \end{matrix} $$ At this point it suffices to verify the map $\wedge^i \kappa^n \to \wedge^{i-1} \kappa^n \otimes \mathfrak m/\mathfrak m^2$ is injective, which can be done by hand.