Definition of matrix exponential Is there an alternative definition of a matrix exponential so I can use it to prove that $$e^{A}=\sum_{m=0}^{\infty} \frac{1}{m!}(A)^m \;?$$
Thanks a lot in advance!

$$\begin{align}&\lim_{n \to \infty} \left(I + \frac{1}{n} A \right)^n = \\ &\lim_{n \to \infty} \left(I^n+(n)I^{n-1}\frac{A}{n}+\frac{(n)(n-1)}{2}I^{n-2}\frac{A^2}{n^2} + \frac{(n)(n-1)(n-2)}{6}I^{n-3}\frac{A^3}{n^3} +\;...\right) = \\ &\lim_{n \to \infty}\left(I + A+\frac{(n)(n-1)}{2n^2}A^2 + \frac{(n)(n-1)(n-2)}{6n^3}A^3 ...\right)= \\ &\lim_{n \to \infty} I + \lim_{n \to \infty}A + \lim_{n \to \infty}\frac{(n)(n-1)}{2n^2}A^2 + \lim_{n \to \infty}\frac{(n)(n-1)(n-2)}{6n^3}A^3 +\; ...\; = \\ &I + \frac{A}{1!} + \frac{A^2}{2!} + \frac{A^3}{3!} +\; ... \;= \\ &\sum_{m=0}^\infty\frac{A^m}{m!} = e^A\end{align}$$
Pretty sure that binomial expansion for matrices is valid when the two matrices commute so this is valid because $IA = AI.$
