= Solution
The assumed <differentiability> makes $e$ continuous. The identity $e'=e$ then implies inductively that $e$ has continuous <derivatives> of every order and
$$
e^{(k)}(t)=e(t)\quad(k\geq0),\qquad e^{(k)}(0)=1.
$$
Fix $x\ne0$. 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 that
$$
e(x)=\sum_{k=0}^n\frac{x^k}{k!}
+\frac{e(\xi_n)}{(n+1)!}x^{n+1},
\qquad
\left|e(x)-\sum_{k=0}^n\frac{x^k}{k!}\right|
\leq M_x\frac{|x|^{n+1}}{(n+1)!}.
$$
For 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 $|x|^{n+1}/(n+1)!\to0$. The <partial sums> converge to $e(x)$, so
$$
\boxed{e(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}\quad\text{for every }x\in\mathbb R.}
$$
At $x=0$ 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.
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