Prove $f(x)=g(x)$ for all $x \in\mathbb{R}$ If 
$$f(x)=\sum_{n=0}^\infty\frac{x^n}{n!}, x\in\mathbb{R}$$
and
$$
g(x) = 1 + \int_0^x f(t) \,dt
$$
prove that $g(x)=f(x)$ for all $x\in\mathbb{R}$
and prove that $f$ is differentiable on $\mathbb{R}$ as well as show that $f'(x)=f(x)$ for all $x\in\mathbb{R}$.
I know that $f$ is continuous on $\mathbb{R}$ and the series $f$ converges uniformly on $[-r,r]$ for all $r>0$, but I'm unsure of how to approach these questions. Any help is greatly appreciated! 
 A: $$f(x)=\sum_{n=0}^\infty \frac{x^n}{n!}=e^x$$
$$g(x)=1+\int_0^x f(t) dt=1+e^x-1=e^x$$
So, we have $f(x)=g(x)$ and $f'(x)=f(x)$
A: The problem is trivial if we use uniform convergence because then we can do term by term differentiation and get $f'(x) = f(x)$ and by definition of $g(x)$ we get $g'(x) = f(x)$ so that $h'(x) = 0$ where $h(x) = f(x) - g(x)$. Then $h(x)$ is a constant and $$h(x) = h(0) = f(0) - g(0) = 1 - 1 = 0$$ and therefore $f(x) = g(x)$.
However I believe the crux of the problem is to establish it without using term by term differentiation. A simple approach is to show that $f(x + y) = f(x)f(y) $ for all $x, y \in \mathbb{R}$ using multiplication of infinite series. Next we can show that $$\lim_{x \to 0}\frac{f(x) - 1}{x} = 1$$ (without using term by term limits). These two facts will establish that $f'(x) = f(x)$ and hence the job is done (as in the first paragraph).
I am wondering if there is any other approach which shows that $f(x) = g(x)$ (without somehow using $f'(x) = f(x)$).
