On the stochastic definition of $e$ I've read on Wikipedia that one can give a stochastic representation of $e$:

In addition to exact analytical expressions for representation of $e$, there are stochastic techniques for estimating $e$. One such approach begins with an infinite sequence of independent random variables $X_1, X_2,\dots$, drawn from the uniform distribution on $[0, 1]$. Let $V$ be the least number $n$ such that the sum of the first $n$ samples exceeds $1$:
  $$V = \min \left \{ n \mid X_1+X_2+\cdots+X_n > 1 \right \}.$$
  Then the expected value of $V$ is $e$: $\mathbb{E}(V) = e$.

I was wondering how to show (analytically) that $\mathbb{E}(V) = e$. I looked at the references but they seems to deal just with numerical aspects.

We have:
$$\mathbb{E}[V]=\sum_{m=0}^{+\infty}\mathbb{P}[V> m]\tag{1}$$
and:
$$ \mathbb{P}[V> m] = \mathbb{P}[X_1+\ldots+X_m\leq 1]\triangleq A_m.\tag{2} $$
The pdf of $S_m=X_1+\ldots+X_m$ can be computed by multiple convolution1: over the interval $[0,1]$ it is given by $\frac{t^{m-1}}{(m-1)!}$, hence $A_m=\frac{1}{m!}$ and:
$$ \mathbb{E}[V]=\sum_{m\geq 0}\frac{1}{m!}=e $$
as wanted.
1) As an alternative approach, notice that: $$\mathbb{P}[X_1+\ldots+X_m\leq 1]=\mu\left(\left\{(x_1,\ldots,x_m)\in[0,1]^m:x_1+\ldots+x_m\leq 1\right\}\right)=\frac{1}{m!}.$$ 
