How to prove this conclusion of a uniform distribution

Question

$X_1,X_2,$.....are i.i.d standard uniform U(0,1),let $$N:=\min\{n \ge2 | X_n>X_{n-1}\}$$ $$T=\min\{n \ge 1 | X_1+...+X_n>1\}$$

We have conclusions that:

1. $\mathbb{E}[N]=e$
2. $\mathbb{E}[T]=e$

Conclusion 1 is easy to prove, but how about 2?

Answer 1

Here is the proof of conclusion 1: $$\mathbb{P}(N>n)=\mathbb{P}(X_1>X_2>...>X_n)=\frac{1}{n!}$$ $$\mathbb{E}[N]=\sum_{n=1}^{\infty}\mathbb{P}(N>n)=e$$