= Solution
For a nonnegative integer-valued <random variable>, its <expected value> is $\mathbb EN=\sum_{r=0}^\infty rp_r$, allowing the value $+\infty$. Since $r=\sum_{n=1}^r1$, rearranging nonnegative summands gives the <tail-sum formula for expectation>:
$$
\mathbb EN=\sum_{r=0}^\infty\sum_{n=1}^r p_r
=\sum_{n=1}^\infty\sum_{r=n}^\infty p_r
=\boxed{\sum_{n=1}^\infty P(N\geq n).}
$$
No integrability assumption is needed for the rearrangement. The same formula holds for an extended nonnegative integer-valued variable by the <monotone convergence theorem>, using $N=\sum_{n\geq1}\mathbf1_{\{N\geq n\}}$.
For the <first ascent in independent continuous observations>, continuity makes ties probability zero. By <order symmetry of independent random variables>, the first $m$ observations have each of their $m!$ strict orderings with equal probability. For $n\geq2$, $N\geq n$ means exactly that the first $n-1$ observations are decreasing, so
$$
P(N\geq1)=1,\qquad P(N\geq n)=\frac1{(n-1)!}\quad(n\geq2).
$$
Subtracting consecutive tails gives
$$
\boxed{P(N=r)=\frac1{(r-1)!}-\frac1{r!}=\frac{r-1}{r!},\quad r\geq2,}
$$
with probability zero for $r=0,1$. The events $\{N\geq n\}$ decrease to $\{N=\infty\}$; continuity of <probability> under decreasing intersections gives $P(N=\infty)=\lim_n1/(n-1)!=0$. Finally,
$$
\boxed{\mathbb EN=1+\sum_{n=2}^\infty\frac1{(n-1)!}=e.}
$$
The first observation cannot itself end the run; this is why the tail includes the separate term $P(N\geq1)=1$.
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