Matrix exponential respects similarity ($e^{PA{P}^{-1}}=P{e}^A{P}^{-1}$)

For an invertible matrix $P$, induction gives $(PA{P}^{-1}{)}^k=P{A}^k{P}^{-1}$, including $k=0$. The power series for the matrix exponential is absolutely convergent in every matrix norm, so multiplication by the fixed matrices $P,P^{-1}$ can be taken through the sum. This proves the displayed identity without any diagonalizability assumption. In particular, if $A=PD{P}^{-1}$ with $D$ diagonal, then $e^A=P\mathrm{diag}(e^{D_{11}},\dots ,e^{D_{nn}})P^{-1}$.