Verify the spectral radius $r(A) = \lim_{n\rightarrow\infty}||A^n||^{1/n}$. I want to Verify the spectral radius $r(A) = \lim_{n\rightarrow\infty}\|A^n\|^{1/n}$. 
Where $A$ is a matrix. 
I have a proof that involves Jordan Blocks. The proof is long and involved but it not to hard to understand. I am interested in knowing if there is an easier (shorter) proof that does not involve using Jordan Blocks. 
Any help and comments are appreciated. Thank you in advance. 

For an alternative proof, you can use the fact that $A^k\rightarrow 0$ (as $k\rightarrow\infty$) iff $r(A)<1$ (which might also involve Jordan blocks, but can be proved also, e.g., using the Schur decomposition).
Let $\|\cdot\|$ be a matrix norm. First, since $r(A)^k=r(A^k)\leq\|A^k\|$, we have $r(A)\leq\|A^{k}\|^{1/k}$. Next, let $\epsilon>0$ and $B=A/(r(A)+\epsilon)$. Since $r(B)=r(A)/(r(A)+\epsilon)<1$, $B^k\rightarrow 0$ as $k\rightarrow\infty$ and hence there is an $N$ such that $\|B^k\|<1$ for $k\geq N$. This means that $\|A^k\|<(\rho(A)+\epsilon)^k$ (if $k\geq N$). Hence for any $\epsilon>0$, there is an $N$ such that $\rho(A)<\|A^k\|^{1/k}\leq\rho(A)+\epsilon$ for $k\geq N$. Consequently, $\lim_{k\rightarrow\infty}\|A^k\|^{1/k}$ exists and is equal to $\rho(A)$.
