= Solution
Induction shows $(PAP^{-1})^k=PA^kP^{-1}$ for every nonnegative integer $k$: in a product, adjacent factors $P^{-1}P$ cancel. At $k=0$, this is $I=PIP^{-1}$. Hence each partial sum of the <matrix exponential> satisfies
$$
\sum_{k=0}^N\frac{(PAP^{-1})^k}{k!}
=P\left(\sum_{k=0}^N\frac{A^k}{k!}\right)P^{-1}.
$$
The series is absolutely convergent, since a submultiplicative <matrix norm> bounds it by $\sum_k\|A\|^k/k!=e^{\|A\|}$. Passing to the limit therefore proves that the <matrix exponential respects similarity>:
$$
\boxed{e^{PAP^{-1}}=Pe^AP^{-1}}.
$$
No <diagonalizability> assumption is required.
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