Integral expansion help! So I'm very close to finishing a proof of the exponential function in terms of differential equations. For this next step, I have to show the following.
For $n \ge 0$ define $E_n (t)$ recursively according to
$$\begin{gather}
E_0 (t) = 1, \\
E_{n+1}(t) = 1 + \int _0 ^t E_n(s)ds.
\end{gather}$$
Show the following: $\displaystyle E(t)= \sum_{n=0}^{+\infty} \frac{t^n}{n!}$.
Any help would be appreciated!
 A: We can do the problem by using Laplace transformation and also convolution of two functions. 
As you noted, the sequence $E_{n+1}(t)=1+\int_0^{t}E_n(k)dk$ is convergent. So if $E_n(t)\to E(t)$ when $n\to\infty$ then we can have: $$E(t)=1+\int_0^{t}E(k)dk$$ According to what we knoe from convolution of two functions, we can write the latter identity as: $$E(t)=1+\int_0^{t}E(k)dk=1+1*E(t)$$ so we are led to the following ODE: $$E(t)=1+1*E(t)$$ Now take  the Laplace of both sides and let $L(E(t))=Y$:  $$Y=L(1+1*E(t))=\frac{1}{s}+\frac{1}{s}\times Y$$ So $Y=\frac{1}{s-1}$. This means that $E(t)=L^{-1}(\frac{1}{s-1})=\exp(t)$ . 
Edit: Or, again we have : $$E(t)=1+\int_0^{t}E(k)dk$$ Differential from both sides, so: $$E'(t)=E(t)$$ a first order differential equation. It is separable and by solving it, you get $$\frac{d(E(t))}{E(t)}=dt$$ and so we have $E(t)=\exp(t)+C$. If you satisfy te condition you were given you will get $c=0$.
A: We could use induction: For $E_1(t)$ we have $E_{1}(t)=1+t=\sum\limits_{n=0}^{1}\frac{t^n}{n!}$. Now assume that for some positive integer $k$ we have
$$E_k(t)=\sum\limits_{n=0}^{k}\frac{t^n}{n!}$$
Then $$E_{k+1}(t)=1+\int\limits_{0}^{t} \sum\limits_{n=0}^{k}\frac{s^n}{n!}ds=1+ 1+\sum\limits_{n=0}^{k}\int\limits_{0}^{t} \frac{s^n}{n!}ds=1+\sum\limits_{n=0}^{k}\frac{t^{n+1}}{(n+1)n!}=1+\sum\limits_{n=1}^{k+1}\frac{t^n}{n!}=\sum\limits_{n=0}^{k+1}\frac{t^n}{n!}$$
By induction we now have $E_{j}(t)=\sum\limits_{n=0}^{j}\frac{t^n}{n!}$ and thus $E(t)=\sum\limits_{n=0}^{\infty}\frac{t^n}{n!}$.
