TITLE: Probability that sum of independent uniform variables is less than 1
QUESTION [5 upvotes]: I would like to determine the probability $\mathbb{P}(X_1+\dots+X_n\leq 1)$, where $X=(X_i)_{1\leq i\leq n}$ is a family of independent uniform random variables on $[0,1]$. My first idea is to do this by induction. The first three base cases are straightforward to determine and give us $\mathbb{P}(X_1\leq 1)=1$, $\mathbb{P}(X_1+X_2\leq 1)=\frac{1}{2}$ and $\mathbb{P}(X_1+X_2+X_3\leq 1)=\frac{1}{6}$, which suggests that $\mathbb{P}(X_1+\dots+X_n\leq 1)=\frac{1}{n!}$. Supposing this is true for a certain arbitrary integer $n$, I am having difficulties establishing the result for $n+1$, i.e. $\mathbb{P}(X_1+\dots+X_n+X_{n+1}\leq 1)=\frac{1}{(n+1)!}$. I believe the starting point should be:
$$\mathbb{P}(X_1+\dots+X_n+X_{n+1}\leq 1)=\mathbb{P}(X_1+\dots+X_n\leq 1-X_{n+1}),$$
and then somehow condition on $X_{n+1}$, but I am stuck at this point of the calculation. Any ideas of references to literature or even an alternative direct proof would be greatly appreciated.

REPLY [5 votes]: A geometric argument should suffice.   Given that $\{X_k\}_\infty$ are all iid Uniform$(0;1)$ random variables, then: 
$\mathsf P(X_1+X_2\leq 1)$ is the probability that points distributed uniformly over the unit square lie in the lower left triangle; which is $1/2$ the area of the unit square.
$\mathsf P(X_1+X_2+X_3\leq 1)$ is the probability that points distributed uniformly over the unit cube lie in the $(0,0,0)$-corner pyramid; which is $1/6$ the volume of the unit cube.
$\mathsf P(X_1+X_2+X_3+X_4\leq 1)$ is the probability that points distributed uniformly over the unit tesseract lie in $(0,0,0,0)$-corner pentachron; which is $1/24$ of the hypervolume of the unit tesseract.
And so forth.
$\mathsf P(\sum\limits_{k=1}^n X_k\leq 1)$ is the probability that points distributed uniformly over a unit $n$-hypercube lie in a corner $n$-hyperpyramid; which is $1/n!$ of the $n$-hypervolume of the unit $n$-hypercube.