Q:

Rolling a die with n sides to get a cumulative score of n

I was told this problem a while ago, and recently someone explained the answer to me, which I didn't understand; could someone please explain in layman's terms (ish)?
You have a die with $n$ sides. Each side is numbered - uniquely - from $1$ to $n$, and has an equal probability of landing on top as the other sides (i.e. a fair die). For large $n$ (I was given it with $n = 1,000,000$), on average how many rolls does it take to achieve a cumulative score of $n$ (or greater)? That is, when you roll it, you add the result to your total score, then keep rolling and adding, and you stop when your score exceeds or is equal to $n$.
The cool thing about this problem: apparently, the answer is $e$. I would like to know exactly how this is derived.

A:

If we divide every roll by $n$, rolling the die and dividing by $n$ approximates the uniform distribution on $[0,1]$ for arbitrarily large $n$.  We then are looking for the expected number of samples from a uniform distribution required to get a sum above $1$.  
For any integer $k \in \mathbb{N}$, Let $X_1, \ldots , X_k, \ldots$ be the random variables in question.  Then 
$$
\mathbb{P}[X_1 + \ldots + X_k \leq 1] = \frac{1}{k!}
$$
as $\frac{1}{k!}$ is the volume of the $k$-dimensional simplex defined by $X_1 + \ldots + X_k \leq 1$.  Now, let $p_k$ be the probability that it takes exactly $k$ rolls to get above $1$; this is equal to 
$$
p_k = \mathbb{P}[X_1 + \ldots + X_{k-1} \leq 1] - \mathbb{P}[X_1 + \ldots + X_{k} \leq 1]  = \frac{1}{(k-1)!} - \frac{1}{k!},$$
as we need the first $k-1$ samples to sum below $1$, and need the first $k$ to sum above $1$.  Thus, the expected number of rolls required is \begin{align}
\mathbb{E}[\text{Number of Rolls Required}] &= \sum\limits_{k = 1}^\infty k\cdot p_k \\
&= \sum\limits_{k = 1}^\infty k \left( \frac{1}{(k-1)!} - \frac{1}{k!}\right) \\
&= \sum\limits_{k = 2}^\infty k \left( \frac{1}{(k-1)!} - \frac{1}{k!}\right) &\text{ as the }k=1\text{ case contributes }0 \\
&= \sum\limits_{k = 2}^\infty \frac{1}{(k - 2)!}\\
&= e.
\end{align}

A:

No real answer, but a way to approach the problem and also to find solutions for small $n$.
Let $\mu_{k}$ denote the expectation of the
number of rolls needed to arrive at a score of at least $k$.
This for $k\in\mathbb{Z}$.
Then:

$\mu_{k}=0$ if $k\leq0$
$\mu_{k}=\frac{1}{n}\left(1+\mu_{k-1}\right)+\frac{1}{n}\left(1+\mu_{k-2}\right)+\cdots+\frac{1}{n}\left(1+\mu_{k-n}\right)=1+\frac{1}{n}\sum_{i=1}^{n}\mu_{k-i}$
if $k>0$.

To be found is $\mu_n$ and we have a recurrence relation.
Edit:
Based on this it can be shown that $\mu_k=(1+\frac1{n})^{k-1}$ for $0<k\leq n$, hence $\mu_n=(1+\frac1{n})^{n-1}\rightarrow e$. See the comments on this answer for that. 
Credit for that goes to Byron, and Marcus spared me some work too.

