TITLE: Is this proof that *on average it takes $e$ straws to break the camel's back* sufficient and clear?
QUESTION [6 upvotes]: Problem: There is a camel with a tolerance of weight 1. We stack straws on his back one by one, and the weight of each straw is a number randomly sampled from a uniform distribution between 0 and 1. Prove that, in expectation, it takes $e$ straws to break the camel's back.

Proof:
Call $P(k, n)$ the probability of surpassing $k$ weight with exactly $n$ straws (before the $n$th straw the total weight was below $k$, and after the $n$th straw the total weight was above $k$). By definition, given that the first straw weighed $x$, the probability of surpassing $k$ weight with the remaining $n-1$ straws is $P(k-x, n-1)$. So if you consider all possible first straw weights $0$ to $k$ (assume $n\geq2$ so that we only want to include the case where the first straw weight is less than $k$), that leads us to
$$P(k, n) = \int_0^k P(k-x, n-1) \mathrm{d}x \quad if n\ge2 \quad (1)$$
$$P(k, 1) = 1-k \quad if n=1 \quad (2)$$
The case of $n=1$ is because we are sampling uniformly from $0$ to $1$ and to surpass $k$ with $1$ straw the sample needs to come from that portion beyond $k$, which is between $k$ and $1$, thus $1-k$.
By definition of expected value, our final answer will be the  infinite weighted sum of probabilities, with the weights being the number of straws:
$$\sum_{m=1}^\infty mP(1, m) \quad (3)$$
So now let's prove the value of $P(k, n)$ using induction.
Induction step: Assume
$$P(k, n) = \frac{k^{(n-1)}(n-k)}{n!} \quad (4)$$
and use that assumption to prove that
$$P(k, n+1) = \frac{k^n(n+1-k)}{(n+1)!} \quad (5)$$
Base case: Prove that the (4) holds for $n=1$.
Induction Step:
$$P(k, n+1) = \int_0^k P(k-x, n) \mathrm{d}x \quad by (1)$$
$$= \int_0^k \frac{(k-x)^{(n-1)}(n-(k-x))}{n!} \mathrm{d}x \quad by (4)$$
$$= \frac{1}{n!}\Big[n \int_0^k (k-x)^{(n-1)} \mathrm{d}x - \int_0^k (k-x)^n \mathrm{d}x\Big]$$
$$= \frac{1}{n!}\Big[\frac{n(-(k-x)^n)}{n}\Big|_{x=0}^k - \frac{-(k-x)^{n+1}}{n+1}\Big|_{x=0}^k\Big]$$
$$= \frac{1}{n!}\Big[k^n - \frac{k^{n+1}}{n+1}\Big]$$
$$= \frac{k^n(n+1-k)}{(n+1)!}$$
Base Case:
$$P(k, 1) = 1 - k$$
$$= \frac{k^{1-1}(1-k)}{1!}$$
Thus we have proved
$$P(k, n) = \frac{k^{(n-1)}(n-k)}{n!} \quad (4)$$
and substituting in $k = 1$ we get
$$P(1, n) = \frac{(n-1)}{n!}$$
Finally, substituting into (3) we get
$$\sum_{m=1}^\infty mP(1, m) = \sum_{m=1}^{\infty} \frac{m(m-1)}{m!} = e$$
$\square$

REPLY [3 votes]: We give an alternate derivation using indicator random variables. Let $X_k=1$ if the sum of the first $k$ weights is $\le 1$, and let $X_k=0$ otherwise. We want $E(1+\sum X_k)$, which is $1+\sum E(X_k)$. 
The probability that the sum of the first $k$ straws is $\le 1$ is the proportion of the unit hypercube of dimension $k$ occupied by the part with $x_1+x_2+\cdots +x_k\le 1$. This proportion is $\frac{1}{k!}$. Thus $\sum E(X_k)=e-1$, and therefore the required expectation is $e$.
Remark: Congratulations on pushing successfully through a difficult calculation. There is overlap between our calculations. I cheated by quoting a standard volume result, while a significant part of your work in effect was an evaluation of the multiple integral that gives us the volume of the region where $x_1+\cdots +x_k\le 1$, with the $x_i$ non-negative.