 Solution

The assumed differentiability makes $e$ continuous. The identity $e^{\prime }=e$ then implies inductively that $e$ has continuous derivatives of every order andFix $x\ne 0$. By the extreme value theorem, the continuous function $|e|$ is bounded on the closed interval between $0$ and $x$; let $M_x$ be its maximum. Applying the Taylor theorem with Lagrange remainder about $0$ gives some ${\xi }_n$ between $0$ and $x$ such thatFor the last factor, successive ratios are $|x|/(n+2)$, which are at most $1/2$ once $n$ is sufficiently large. A geometric sequence bound therefore proves that $|x{|}^{n+1}/(n+1)!\to 0$. The partial sums converge to $e(x)$, soAt $x=0$ the equality follows directly from $e(0)=1$. The proof obtains this Taylor series solely from the differential identity, continuity on a fixed finite interval, and the Taylor remainder estimate.