Waiting for a decreasing run to break

Let $X_1, X_2, \dots$ be i.i.d. draws from a continuous distribution. Define $N$ as the first index at which the sequence goes up, i.e. $X_1\ge X_2\ge\dots\ge X_{N-1}$ and $X_{N-1}<X_N$. Compute $\ln\left(\mathbb{E}[N]\right)$.

$1$. $N\ge k$ means the first $k-1$ values are in decreasing order, which has probability $\tfrac{1}{(k-1)!}$ since all orderings are equally likely. Summing tail probabilities, $\mathbb{E}[N]=\sum_{k\ge1}\tfrac1{(k-1)!}=e$, so the log is $1$.