Why does the number of continuous uniform variables on (0,1) needed for their sum to exceed one have mean $e$? Let us sum a stream of random variables, $X_i \overset{iid}\sim \mathcal{U}(0,1)$; let $Y$ be the number of terms we need for the total to exceed one, i.e. $Y$ is the smallest number such that
$$X_1 + X_2 + \dots + X_Y > 1.$$
Why does the mean of $Y$ equal Euler's constant $e$?
$$\mathbb{E}(Y) = e = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots $$
 A: In Sheldon Ross' A First Course in Probability there is an easy to follow proof:
Modifying a bit the notation in the OP, $U_i \overset{iid}\sim \mathcal{U}(0,1)$ and $Y$ the minimum number of terms for $U_1 + U_2 + \dots + U_Y > 1$, or expressed differently:
$$Y = min\Big\{n: \sum_{i=1}^n U_i>1\Big\}$$
If instead we looked for:
$$Y(u) = min\Big\{n: \sum_{i=1}^n U_i>u\Big\}$$ for $u\in[0,1]$, we define the $f(u)=\mathbb E[Y(u)]$, expressing the expectation for the number of realizations of uniform draws that will exceed $u$ when added.
We can apply the following general properties for continuous variables:
$E[X] = E[E[X|Y]]=\displaystyle\int_{-\infty}^{\infty}E[X|Y=y]\,f_Y(y)\,dy$
to express $f(u)$ conditionally on the outcome of the first uniform, and getting a manageable equation thanks to the pdf of $X \sim U(0,1)$, $f_Y(y)=1.$ This would be it:
$$f(u)=\displaystyle\int_0^1 \mathbb E[Y(u)|U_1=x]\,dx \tag 1$$
If the $U_1=x$ we are conditioning on is greater than $u$, i.e. $x>u$, $\mathbb E[Y(u)|U_1=x] =1 .$ If, on the other hand, $x <u$, $\mathbb E[Y(u)|U_1=x] =1 + f(u - x)$, because we already have drawn $1$ uniform random, and we still have the difference between $x$ and $u$ to cover. Going back to equation (1):
$$f(u) = 1 + \displaystyle\int_0^x f(u - x) \,dx$$, and with substituting $w = u - x$ we would have $f(u) = 1 + \displaystyle\int_0^x f(w) \,dw$.
If we differentiate both sides of this equation, we can see that:
$$f'(u) = f(u)\implies \frac{f'(u)}{f(u)}=1$$
with one last integration we get:
$$log[f(u)] = u + c \implies f(u) = k \,e^u$$
We know that the expectation that drawing a sample from the uniform distribution and surpassing $0$ is $1$, or $f(0) = 1$. Hence, $k = 1$, and $f(u)=e^u$. Therefore $f(1) = e.$
A: First observation: $Y$ has a more pleasing CDF than PMF
The probability mass function $p_Y(n)$ is the probability that $n$ is "only just enough" for the total to exceed unity, i.e. $X_1 + X_2 + \dots X_n$ exceeds one while $X_1 + \dots + X_{n-1}$ does not.
The cumulative distribution $F_Y(n) = \Pr(Y \leq n)$ simply requires $n$ is "enough", i.e. $\sum_{i=1}^{n}X_i > 1$ with no restriction on how much by. This looks like a much simpler event to deal with the probability of.
Second observation: $Y$ takes non-negative integer values so $\mathbb{E}(Y)$ can be written in terms of the CDF
Clearly $Y$ can only take values in $\{0, 1, 2, \dots\}$, so we can write its mean in terms of the complementary CDF, $\bar F_Y$.
$$\mathbb{E}(Y) = \sum_{n=0}^\infty \bar F_Y(n) = \sum_{n=0}^\infty \left(1 - F_Y(n) \right)$$ 
In fact $\Pr(Y=0)$ and $\Pr(Y=1)$ are both zero, so the first two terms are $\mathbb{E}(Y) = 1 + 1 + \dots$.
As for the later terms, if $F_Y(n)$ is the probability that $\sum_{i=1}^{n}X_i > 1$, what event is $\bar F_Y(n)$ the probability of?
Third observation: the (hyper)volume of an $n$-simplex is $\frac{1}{n!}$
The $n$-simplex I have in mind occupies the volume under a standard unit $(n-1)$-simplex in the all-positive orthant of $\mathbb{R}^n$: it is the convex hull of $(n+1)$ vertices, in particular the origin plus the vertices of the unit $(n-1)$-simplex at $(1, 0, 0, \dots)$, $(0, 1, 0, \dots)$ etc.

For example, the 2-simplex above with $x_1 + x_2 \leq 1$ has area $\frac{1}{2}$ and the 3-simplex with $x_1 + x_2 + x_3 \leq 1$ has volume $\frac{1}{6}$.
For a proof that proceeds by directly evaluating an integral for the probability of the event described by $\bar F_Y(n)$, and links to two other arguments, see this Math SE thread. The related thread may also be of interest: Is there a relationship between $e$ and the sum of $n$-simplexes volumes?
A: Fix $n \ge 1$.  Let $$U_i = X_1 + X_2 + \cdots + X_i \mod 1$$ be the fractional parts of the partial sums for $i=1,2,\ldots, n$.  The independent uniformity of $X_1$ and $X_{i+1}$ guarantee that $U_{i+1}$ is just as likely to exceed $U_i$ as it is to be less than it.  This implies that all $n!$ orderings of the sequence $(U_i)$ are equally likely.
Given the sequence $U_1, U_2, \ldots, U_n$, we can recover the sequence $X_1, X_2, \ldots, X_n$.  To see how, notice that


*

*$U_1 = X_1$ because both are between $0$ and $1$.

*If $U_{i+1} \ge U_i$, then $X_{i+1} = U_{i+1} - U_i$.

*Otherwise, $U_i + X_{i+1} \gt 1$, whence $X_{i+1} = U_{i+1} - U_i + 1$.
There is exactly one sequence in which the $U_i$ are already in increasing order, in which case $1 \gt U_n = X_1 + X_2 + \cdots + X_n$.  Being one of $n!$ equally likely sequences, this has a chance $1/n!$ of occurring.  In all the other sequences at least one step from $U_i$ to $U_{i+1}$ is out of order.  This implies the sum of the $X_i$ had to equal or exceed $1$.  Thus we see that
$$\Pr(Y \gt n) = \Pr(X_1 + X_2 + \cdots + X_n \le 1) =  \Pr(X_1 + X_2 + \cdots + X_n \lt 1) = \frac{1}{n!}.$$
This yields the probabilities for the entire distribution of $Y$, since for integral $n\ge 1$
$$\Pr(Y = n) = \Pr(Y \gt n-1) - \Pr(Y \gt n) = \frac{1}{(n-1)!} - \frac{1}{n!} = \frac{n-1}{n!}.$$
Moreover,
$$\mathbb{E}(Y) = \sum_{n=0}^\infty \Pr(Y \gt n) = \sum_{n=0}^\infty \frac{1}{n!} = e,$$
QED.
