Density of the sum of $n$ uniform(0,1) distributed random variables I am working on the following problem:

Let $X_1, X_2, \ldots, X_n, \ldots$ be iid. random variables, each of Uniform$(0,1)$ distribution. Denote by $f_n(x)$ the density of the random variable $S_n := \sum_{k = 1}^n X_k$. Then 
  \begin{align*}
f_n(x) = \frac{1}{(n-1)!} \sum_{k = 0}^{[x]} (-1)^k \binom{n}{k} (x-k)^{n-1},
\end{align*}
  where $[x]$ denotes the floor function. 

I tried this, but I think something went wrong:
Prove by induction. For $n = 1$
\begin{align*}
f_1(x) = \frac{1}{(1-1)!} \sum_{k = 0}^{[x]} (-1)^k \binom{1}{k} (x-k)^{1-1} &= \sum_{k = 0}^{[x]} (-1)^k \binom{1}{k}
=\begin{cases}
1, & 0 \le x < 1 \\
0, & \text{otherwise}
\end{cases}
\end{align*}
Furthermore
\begin{align*}
f_{n+1}(x) &= \mathbb P(S_{n+1} = x) = \mathbb P(S_n + X_{n+1} = x) \\
&= \sum_{m = 0}^{\infty} \mathbb P(S_n + X_{n+1} = x \mid X_{n+1} = m) \mathbb P(X_{n+1} = m) \\
&= \sum_{m = 0}^{\infty} \mathbb P(S_n + m = x)  \cdot 1_{[0,1]}(m) 
= \mathbb P(S_n = x) + \mathbb P(S_n = x-1) \\
&= \frac{1}{(n-1)!} \sum_{k = 0}^{[x]} (-1)^k\binom{n}{k}(x-k)^{n-1} + \frac{1}{(n-1)!} \sum_{k = 0}^{[x-1]} (-1)^k\binom{n}{k}(x-1-k)^{n-1} \\
&=  \frac{1}{(n-1)!}\left(\sum_{k = 0}^{[x]} (-1)^k\binom{n}{k}(x-k)^{n-1} + \sum_{k = 0}^{[x]-1} (-1)^k\binom{n}{k}(x-1-k)^{n-1}\right) \\
&=  \frac{1}{(n-1)!}\left(x^{n-1}+\sum_{k = 1}^{[x]} (-1)^k\binom{n}{k}(x-k)^{n-1} + \sum_{k = 1}^{[x]} (-1)^{k-1}\binom{n}{k-1}(x-1-(k-1))^{n-1}\right) \\
&=  \frac{1}{(n-1)!}\left(x^{n-1}+\sum_{k = 1}^{[x]} (-1)^k\binom{n}{k}(x-k)^{n-1} - (-1)^{k}\binom{n}{k-1}(x-k)^{n-1}\right).
\end{align*}
How can I solve this?
Edit:
\begin{align*}
f_{n+1}(x) &= (f_n * f_{X_{n+1}})(x) 
= \int_{-\infty}^\infty f_{n}(y) f_{X_{n+1}}(x-y) \, dy \\
&= \int_{-\infty}^\infty \frac{1}{(n-1)!} \sum_{k = 0}^{[y]} (-1)^k \binom{n}{k}(y-k)^{n-1} \cdot 1_{(0,1)}(x-y) \, dy \\
&= \frac{1}{(n-1)!} \int_{-\infty}^{\infty} \sum_{k = 0}^{[y]} (-1)^k \binom{n}{k}(y-k)^{n-1} \cdot 1_{(x-1, x)}(y)\, dy \\
&= \frac{1}{(n-1)!} \int_{x-1}^{x} \sum_{k = 0}^{[y]} (-1)^k \binom{n}{k}(y-k)^{n-1} \, dy
\end{align*}
I want to swap the sum and the integral, but the sum depends on $y$.

Here is a totally different method.  So you want to compute $I*I*\cdots*I = I^{*n}$, where $I$ is the indicator function of $[0,1]$, and the $*$ denotes convolution, and the superscript $I^{*n}$ denotes the $n$th convolution with itself.
Differentiate $n$ times.  You will get $(\delta_1 - \delta_0)^{*n}$, where $\delta_x$ denotes the Dirac delta function centered at $x$.  This is easily expanded using the Binomial Theorem, remembering that $\delta_x * \delta_y = \delta_{x+y}$.
Now take the indefinite integral $n$ times.  The tricky bit is how to consider the constants of integration.  But the easiest boundary condition is to suppose that all the derivatives disappear at $-\infty$.  In this way $\int^n \delta_y (dx)^n = \frac1{(n-1)!} (x-y)_+^{n-1}$, where $x_+$ denotes $\max\{x,0\}$.  (Please excuse the notation for $n$th indefinite integral.)
I got this method from an exercise in the book by Nakhle Asmar on Partial Differential Equations.
