I've been working on solving some problem and reached to the following equation: $\sum_{k=0}^\infty \big[\frac{X^k}{k!} + \frac{kX^{k-1}}{k!} \big] = \sum_{k=0}^\infty \big[\frac{X^k}{k!} + \frac{X^{k-1}}{(k-1)!} \big]$ ($X$ is a complex matrix, but not too important for the purpose of this equation).
I am wondering whether the second term starts from $k = 1$ since for $k = 0$ we have $-1!$ in the denominator, but I am not entirely sure if this is right. If my guess is wrong, then I'd like to know how the second term at $k=0$ should be calculated.
In this simplification, you cancelled out a $k$ in the numerator with one in the denominator. But when $k = 0$, there is no $k$ in the denominator. $0! = 1$ by definition. It is not $0$ times anything. Thus you did a nonsense operation and got a nonsense result. The calculation you needed is $$\begin{align}\sum_{k=0}^\infty \left[\frac{X^k}{k!} + \frac{kX^{k-1}}{k!} \right] &= \sum_{k=0}^\infty \frac{X^k}{k!} + \sum_{k=0}^\infty\frac{kX^{k-1}}{k!} \\ &= e^X + \left(0 + \sum_{k=1}^\infty\frac{kX^{k-1}}{k!}\right)\\ &= e^X + \sum_{k=1}^\infty\frac{X^{k-1}}{k-1!}\\ &= e^X + \sum_{n=0}^\infty\frac{X^n}{n!}\\ &= 2e^X\end{align}$$
Now, one could object to saying that $\frac{0X^{-1}}{0!} = 0$, since it quite possible that $X^{-1}$ is not even defined, but if so, then the problem already exists in the original expression. And almost certainly, however you obtained that expression, the term should be interpreted as $0$ even if $X^{-1}$ does not exist. But without knowing that derivation, I cannot be sure.