Function as a series : Let $f(x)=\sum_{n=0}^{+\infty}\dfrac{x^n}{n!}$. Verify that
$$\int_0^xf(x)dt=f(x)-1$$
This is the exercise 3 of the section $7.4$, of Guidorizzi's Calculus, Vol. 4.
What I have tried:
By the ratio test, $\sum_{n=0}^{+\infty}\frac{x^n}{n!}$ converges. So, $\forall x\in\Bbb{R}$, $f(x)=\sum_{n=0}^{+\infty}\frac{x^n}{n!}$ is a real (constant) number. Then
$$\int_0^xf(x)dt=[f(t)t]_0^x=xf(x)-0f(x)=xf(x)$$
So I have tried to make some computation, but I cannot see how $xf(x)$ could be equal to $f(x)-1$. 
 A: First way:
$$f(x)=\sum_{n=0}^\infty \frac{x^n}{n!}=e^x.$$
Then, $$\int_0^x f(t)dt=\int_0^x e^t dt=\left[e^t\right]_0^x=e^x-e^0=e^x-1=f(x)-1.$$
Second way:
$$\sum_{n=0}^\infty \frac{x^n}{n!}$$
converge uniformly (to $e^x$, but this is not the problem), then you can permute limits and integral. Therefore,
$$\int_0^x f(t)dt=\int_0^x \sum_{n=0}^\infty \frac{t^n}{n!}dt=\sum_{n=0}^\infty \int_0^x \frac{t^n}{n!}dt=\sum_{n=0}^\infty \frac{x^{n+1}}{(n+1)!}=\sum_{n=1}^\infty \frac{x^n}{n!}=\sum_{n=0}^\infty \frac{x^n}{n!}-1=f(x)-1.$$
Q.E.D.
A: You should use the fact that the sum is uniform convergent in the interval and substitute the sum and integration.
A: You are misunderstanding power series. The function $f$ is not constant. 
What you have to do is integrate term by term, apply the fundamental theorem of calculus and rewrite the resulting series conveniently in order to transform the expression in something in terms of $f$.
A: We have $$f(x) = \sum\limits_{n=0}^\infty \frac{x^n}{n!} $$
Thus $$ \int\limits_0^x f(t) dt = \int\limits_0^x \sum\limits_{n=0}^\infty \frac{t^n}{n!} = \sum\limits_{n=0}^\infty \int\limits_0^x\frac{t^n}{n!}.$$
Can you take it from here?
