<s>[INST] Differentiation of Power Series Let $$f(x)= \sum_{n=0}^{\infty} \frac{x^n}{n!} $$
for $x\in \mathbb R$. 
Show $f′ =f$.
Note: Do not use the fact that $f(x) = e^x$. This is true but has not been established at this point in the text.
 [/INST] Here is a proof which avoids uniform convergence.
Let $$f(x) = \sum_{n = 0}^{\infty}\frac{x^{n}}{n!}\tag{1}$$ Using multiplication of infinite series it is easy to show that $$f(x)f(y) = f(x + y)\tag{2}$$ for all real $x, y$ (the proof of this also requires binomial theorem for positive integer index).
Next we can see that if $0 < x < 2$ then we have 
\begin{align}
\frac{f(x) - 1}{x} &= 1 + \frac{x}{2!} + \frac{x^{2}}{3!} + \cdots\notag\\
&\leq 1 + \frac{x}{2} + \frac{x^{2}}{2^{2}} + \frac{x^{3}}{2^{3}} + \cdots\notag\\
&= 1 + \frac{x/2}{1 - (x/2)} = 1 + \frac{x}{2 - x}\notag
\end{align}
Thus for $0 < x < 2$ we have $$1 \leq \frac{f(x) - 1}{x}\leq 1 + \frac{x}{2 - x}\tag{3}$$ and taking limits when $x \to 0^{+}$ and using Squeeze theorem we get $$\lim_{x \to 0^{+}}\frac{f(x) - 1}{x} = 1\tag{4}$$ This also shows that $$\lim_{x \to 0^{+}}f(x) = \lim_{x \to 0^{+}}1 + x\cdot\frac{f(x) - 1}{x} = 1 + 0\cdot 1 = 1\tag{5}$$ Again using equation $(2)$ we have $$f(x)f(-x) = f(x + (-x)) = f(0) = 1$$ so that $f(-x) = 1/f(x)$ and therefore
\begin{align}
\lim_{x \to 0^{-}}\frac{f(x) - 1}{x}&= \lim_{y \to 0^{+}}\frac{1 - f(-y)}{y}\text{ (putting }x = -y)\notag\\
&= \lim_{y \to 0^{+}}\frac{f(y) - 1}{y}\cdot\frac{1}{f(y)}\notag\\
&= 1\cdot 1 = 1\text{ (using equations (4) and (5))}\notag
\end{align}
It now follows that $$\lim_{x \to 0}\frac{f(x) - 1}{x} = 1\tag{6}$$ Finally we have
\begin{align}
f'(x) &= \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}\notag\\
&= \lim_{h \to 0}\frac{f(x)f(h) - f(x)}{h}\notag\\
&= \lim_{h \to 0}f(x)\cdot\frac{f(h) - 1}{h}\notag\\
&= f(x)\cdot 1 = f(x)\text{ (using equation (6))}\notag
\end{align}
</s>