= Solution
Termwise differentiation of the absolutely convergent <matrix exponential> series gives
$$
\frac{d}{dt}e^{tP}=e^{tP}P=Pe^{tP}.
$$
Because $P$ and $Q$ <commute>, $P$ also <commutes> with every power of $Q$ and hence with $e^{tQ}$. The <product rule> therefore gives
$$
f_1'(t)=e^{tP}Pe^{tQ}+e^{tP}e^{tQ}Q=f_1(t)(P+Q).
$$
Direct differentiation gives $f_2'(t)=f_2(t)(P+Q)$, and both initial values are $I$. The <uniqueness theorem for ordinary differential equations>, applied to the finitely many <matrix> entries, implies $f_1=f_2$. One can see uniqueness directly: the difference $D=f_1-f_2$ obeys $D'=D(P+Q)$ and $D(0)=0$, so
$$
\frac{d}{dt}\left(De^{-t(P+Q)}\right)=0
$$
and the product is identically zero. At $t=1$ this yields
$$
\boxed{e^Pe^Q=e^{P+Q}\quad\text{when }PQ=QP.}
$$
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