Showing $e^\Omega = \sum_{n=1}^\infty \frac{\Omega^n}{n!}$ for non-diagonal matrix $\Omega$

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Context. I am currently reading a Quantum Mechanics book in which it is stated that it is "obvious" (without proof) that

$$ e^{(α Ω)} e^{(β Ω)} = e^{((α + β) Ω)} $$

where $\alpha, \beta \in \mathbb{C}$ and $\Omega$ is a square matrix of complex scalars, and $e^{\Omega}$ is defined as

$$ e^\Omega = \sum_{n=\color{red}{0}}^\infty \frac{\Omega^n}{n!} $$

whenever $\lim_{n \to \infty} \Omega^n$ converges.

Problem: I only know how to show this is true when $\Omega$ happens to be a diagonal matrix (in which case the matrix multiplication of $\Omega^n$ devolves to multiplication of the scalars $d_i^n$ along the diagonal of $\Omega$).

How does one show this result holds in general, when $\Omega$ isn't necessarily diagonal?