I'm wondering if
$(e^{A})_{ij} = e^{A_{ij}}$
I should prove that
$(e^{A})_{ij} = \left(\sum\limits_{k = 0}^\infty \frac{A^k}{k!}\right)_{ij} = (\mathbb{I} + A + \frac{1}{2}A^2 + ...)_{ij} = \mathbb{I}_{ij} + A_{ij} + \frac{1}{2}A_{ih}A_{hj} + ... = \delta_{ij} + A_{ij} + \frac{1}{2}A_{ih}A_{hj} + ... \stackrel{?}{=} \sum\limits_{k = 0}^\infty \frac{(A_{ij})^k}{k!} = e^{A_{ij}}$
But I would say that since $\delta_{ij}$ is not always $1$, but only when $i = j$, then what I'd want to prove is false, since the exponential always needs instead the $1$ at the beginning of the expansion.
Can you tell me if I'm right? Or, else, how would you prove/disprove that?