For any $\epsilon>0$ , consider the matrix $$ \tilde{A}=(\rho(A)+\epsilon)^{-1}A $$ Then, obviously, $$ \rho(\tilde{A}) = \frac{\rho(A)}{\rho(A)+\epsilon}<1 $$ and, by a well-known result on convergence of matrix powers, $$ \lim_{k \to \infty}\tilde{A}^k=0. $$ That means, by sequence limit definition, a natural number $N_1\in \mathbf{N}$ exists such that $$ \forall k\geq N_1 \Rightarrow \|\tilde{A}^k\| < 1 $$ which in turn means: $$ \forall k\geq N_1 \Rightarrow \|A^k\| < (\rho(A)+\epsilon)^k $$ or $$ \forall k\geq N_1 \Rightarrow \|A^k\|^{1/k} < (\rho(A)+\epsilon). $$

Let's now consider the matrix $$ \check{A}=(\rho(A)-\epsilon)^{-1}A $$ Then, obviously, $$ \rho(\check{A}) = \frac{\rho(A)}{\rho(A)-\epsilon}>1 $$ and so, by the same convergence theorem,$\|\check{A}^k\|$ is not bounded. This means a natural number $N_2\in \mathbf{N}$ exists such that $$ \forall k\geq N_2 \Rightarrow \|\check{A}^k\|>1 $$ which in turn means: $$ \forall k\geq N_2 \Rightarrow \|A^k\| > (\rho(A)-\epsilon)^k $$ or $$ \forall k\geq N_2 \Rightarrow \|A^k\|^{1/k} > (\rho(A)-\epsilon). $$ Taking $N:=max(N_1,N_2)$ and putting it all together, we obtain: $$ \forall \epsilon>0, \exists N\in\mathbb{N}: \forall k\geq N \Rightarrow \rho(A)-\epsilon < \|A^k\|^{1/k} < \rho(A)+\epsilon $$ which, by definition, is $$ \lim_{k \to \infty}\|A^k\|^{1/k} = \rho(A).\,\,\square $$

Actually, in case the norm is self-consistent, the proof shows more than the thesis; in fact, using the fact that $|\lambda|\leq\rho(A)$ , we can replace in the limit definition the left lower bound with the spectral radius itself and write more precisely: $$ \forall \epsilon>0, \exists N\in\mathbb{N}: \forall k\geq N \Rightarrow \rho(A) \leq \|A^k\|^{1/k} < \rho(A)+\epsilon $$ which, by definition, is $$ \lim_{k \to \infty}\|A^k\|^{1/k} = \rho(A)^+. $$