Computing this expected value

Question

If $$N(x)=\min \left \{ n : \sum_{i=1}^nU_i>x\right \}$$ where the $U_i$ are i.i.d. uniformly distributed on $(0,1)$ and $x\in (0,1)$, compute $\mathbb{E}[N(x)]$ by first proving the following statement by induction: $$\mathbb{P}(N(x)\ge n+1)=\frac {x^n}{n!}$$ It works for $n=0$, next I assume it works for $n-1$ and try proving it for $n$ by conditioning on whether $U_1\le x$ or $U_1 > x$. After noting that $$\mathbb{P}(N(x)\ge n+1|U_1 > x)=0$$ I reach the following expression $$\mathbb{P}(N(x)\ge n+1)=\frac {x(x-u_1)^{n-1}}{(n-1)!}$$ However this does not reduce to $\frac {x^n}{n!}$ no matter how I expand the polynomial and try to get rid of $u_1$. I know of another solution involving differential equations but I want to solve it this way, what am I doing wrong?

Answer 1

$$P(N(x)\ge n+1)=\int_0^x P(N(x)\ge n+1|U_1=t)dt=\int_0^x P(N(x-t)\ge n)dt\\ =\int_0^x\frac{(x-t)^n}{n!}dt=\left.-\frac{(x-t)^{n+1}}{(n+1)!}\right|_{t=0}^{t=x}=\frac{x^{n+1}}{(n+1)!}$$