Numbers are drawn randomly from the set ${1, ..., n}$ until their sum exceeds $k$ for $0 \leq k \leq n$. Show that the expected number of draws equals
$$e_k = \big(1 + \frac{1}{n}\big)^k.$$ (hint: show that $e_k = 1 + \frac{1}{n}[e_0 + ... + e_{k-1}]$.)
Note: It was not specified in the problem but I believe $k$ is an integer and we are drawing with replacement.
Attempt 1: This question is very similar to a putnum problem with solutions explained here. In this approach, I should compute $P(U_1 + ... + U_m \geq k) = 1 - P(U_1 + ... + U_m < k)$ where U is the drawn number. But doing so we would inevitably have to find the volume of a $m$ dimensional simplex with side lengths $k$, which is $\frac{k^m}{m!}.$ Here the $k^m$ term screws up everything when I try to simplify things, while in that classic problem it was simply $\frac{1}{m!}$. Basically, I don't think this works (and even if it does, it definitely isn't pretty).
Attempt 2: If we assume the hint, I can show the desired result by some algebra. Problem is, I'm not sure how to show the hint. Here's my thinking process. After $k-1$ draws, our sum is anywhere between $1$ and $k$, and the next draw we need to pick something that will bring us over $k$ altogether. If our current sum is 1, then the number we need to pick is $\geq k$, whose probability is given by $\frac{n-k}{k}$. Doing for all terms, we have the following recursive formula: $$e_k = (e_0 + 1)\frac{n-k}{n} + (e_1 + 1)\frac{n-k+1}{n} + ...+ (e_{k-1}+1)\frac{n}{n}$$
My question: Does this recursion look right? I'm quite confident but again my algebra doesn't work out to the desired expression: $e_k = 1 + \frac{1}{n}[e_0 + ... e_{k-1}].$ Please let me know what I did wrong.