Uniform-sum crossing time below one ($\mathbb{E}N(x)=e^x(0\le x\le 1)$)

For independent uniform distribution increments on $(0,1)$ and $0\le x\le 1$, let $N(x)$ be the first index at which their partial sum exceeds $x$. First-step analysis gives $M(x)=\mathbb{E}N(x)=1+{\int }_0^{x}M(s)ds$, hence $M=e^x$. Equivalently, $\mathbb{P}(N(x)>n)=x^n/n!$ is a simplex-volume probability and the tail-sum formula sums to $e^x$. The strict inequality gives $N(0)=1$ almost surely.