The probability distribution of the sum of independent uniform random variables in $\left[0,1\right]$ Let me preface this by saying that this is not a question about the Irwin-Hall distribution , at least directly. Rather, in the proof of Lemma 1 in the appendix of this paper , the author shows that the joint density of $S_n:= \sum_{I=1}^{n}X_i$ where $X_i\stackrel{\rm i.i.d}{\sim}U\left[0,1\right]$ is given by $$f_n(s) = \frac{s^{n-1}}{(n-1)!}P\left(\max_{1\leq i\leq n} L_n \leq s^{-1}\right)$$ where $\{L_i\}_{i=1}^n$ are the lengths of consecutive sub-intervals of $\left[0,1\right]$ cut out by $(n-1)$ uniform random numbers in the interval. I know that the the joint density of the first $(n-1)$ such lengths is $(n-1)!$ on its support. However, I failed to follow the argument of the author to its conclusion. 
Specifically, he argues that the joint density of $S_n,X_1,\ldots,X_{n-1}$ is the same as the joint density of the $n$ independent uniform which is 1 on its support. Then, applying a transformation, the joint density of $S_n,\frac{X_1}{S_n},\ldots,\frac{X_{n-1}}{S_n}$ can be found to be $s^{n-1}$. I cannot follow the argument from this point on, where he presumably integrates out the other variables to derive $f(s)$. If anyone could clarify this step for me, it'd be great.
 A: Ok so this answer basically avoids what Pittel does (maybe that is easier but it's not obvious to me), and follows Feller's approach. 
We have $X_1$, $\cdots$, $X_n$ independent uniform random variables on $[0,a]$. Let $S_n$ be their sum. We want the density of $S_n$, let's call that $f_{n,a}$. Obviously $f_{1,a}$ is $\frac{1}{a}$ on $[0,a]$ and $0$ on the rest of the positive real line. And the rest of the $f_{n,a}$ obey the convolution recurrence, so for $n \geq 2$,
$$f_{n,a}(x) = \frac{1}{a} \int_0^a f_{n-1,a}(x-y) \, dy.$$
Let's set this aside and turn to the other item Pittel uses. Say we partition $[0,t]$ into n intervals using independent uniform random points $Y_1, \cdots, Y_{n-1}$. We want to track the probability that none of the subintervals are larger than $a$, i.e. $P(\max_{1 \leq i \leq n-1} \left(Y_{i+1} - Y_i\right) \leq a)$. Call this thing $\varphi_{n,a}(t)$. By this definition, $\varphi_{1,a}(t)$ is clearly $1$ for $0 \leq t \leq a$, and $0$ on the rest of the positive real line.
What about for larger $n$? We have $n-1$ choices for a random point $Y_i$. Once we choose, call its position $x$. Then the probability that the $Y_i$ we chose is the leftmost one is $\left(\frac{t-x}{t}\right)^{n-2}$. 
Also, this leftmost point must be less than $a$.  Now the remaining variables are distributed with the same conditions over $[x,t]$, so the conditional probability they satisfy the requisite conditions is $\varphi_{n-1,a}(t-x)$. Integrate this over the possible $x$ and get
$$\varphi_{n,a}(t) = (n-1) \int_0^a \varphi_{n-1,a}(t-x) \left(\frac{t-x}{t}\right)^{n-2} \, \frac{1}{t}dx.$$
Now set 
$$u_{n,a}(t) = \frac{\varphi_{n,a}(t)t^{n-1}}{a^{n-1}(n-1)!},$$ 
and observe for $n \geq 2$,
$$ u_{n,a}(t) = \frac{1}{a} \int_0^a u_{n-1,a}(t-x) \, dx.$$ 
This matches our recurrence above and the initial condition matches as well when $a=1$. So $u_{n,1}$ is our desired density function. As the problem requires, we have $a=1$, then put $t = s$ to get what Pittel has:
$$ u_{n,1}(s) = \frac{s^{n-1}}{(n-1)!} \varphi_{n,1}(s),$$
where the $\varphi$ term is what you want: the probability that no sub-interval exceeds length $1$ when $[0,s]$ is cut up into $n$ sub-intervals, or after rescaling, the probability that no sub-interval exceeds length $\frac{1}{s}$ when $[0,1]$ is cut up into $n$ sub-intervals.
