Does $(e^a)^b=e^{ab}$ hold for the matrix exponential?

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Here's the textbook definition of the matrix exponential. Let $A$ be a $n\times n$ matrix, then

$$e^A=\sum_{n=0}^{\infty} \dfrac{1}{n!}A^{n}.$$

It's not quite clear to me how to work this series out. Some properties are quite intuitive and resemble the well-known $e^x$, where $x$ is a real number. I'd like to know if, for some constant positive integer $t$,

$$\left(e^A\right)^t=e^{tA}.$$

I've crunched some numbers and it seems to hold, but I have no clue on how to prove it. Is this true?

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As mentioned in the comments, the proof follows from the more general fact: if $A$ and $B$ are commuting matrices, then $e^{A+B}=e^{A}e^{B}$.

Let $t$ be a positive integer and $A$ a $n\times n$ matrix. Obviously, $A$ commutes with itself, thus, by the fact pointed out above,

$$(e^{A})^t=\underbrace{e^A\dots e^A}_\text{t times}=e^{A+\dots +A}=e^{tA}.$$

Therefore, the equality holds.