On the stochastic definition of $e$
Solution 1:
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: $$\mathbb{P}[X_1+\ldots+X_m\leq 1]=\mu\left(\left\{(x_1,\ldots,x_m)\in[0,1]^m:x_1+\ldots+x_m\leq 1\right\}\right)=\frac{1}{m!}.$$