Let $X_1,X_2,\dots,X_n,\dots$ be iid. random variables, each of Uniform$(0,1)$ distribution. Denote by $f_n(x)$ the density of the random varible $S_n := \Sigma^n_{k=1}X_k$. Prove $$ f_n(x)=\frac{1}{(n-1)!}\sum^{\lfloor x \rfloor }_{k=0}(-1)^k{n\choose k}(x-k)^{n-1} $$
My first intuition is to do this by induction on $n$. I decided to try to work out the density $f_{n+1}(x)$ by convolving another Uniform$(0,1)$ with $f_n(x)$. I came up with $$ f_{n+1}(x)=\int_{x-1}^x\frac{1}{(n-1)!}\sum^{\lfloor y \rfloor }_{k=0}(-1)^k{n\choose k}(y-k)^{n-1}dy $$ where the integral is between $x-1$ and $x$ because the density of the Uniform$(0,1)$ is $0$ everywhere except bewteen $0$ and $1$. I wasn't really sure where to go from here, but I then tried to simplify the integral by separating it into the part where $\lfloor y \rfloor$ is equal to $\lfloor x \rfloor$, and the part where $\lfloor y \rfloor$ is equal to $\lfloor x-1 \rfloor$. So $$ f_{n+1}(x)=\int_{x-1}^{\lfloor x\rfloor}\frac{1}{(n-1)!}\sum^{\lfloor x-1 \rfloor }_{k=0}(-1)^k{n\choose k}(y-k)^{n-1}dy + \int_{\lfloor x \rfloor}^{x}\frac{1}{(n-1)!}\sum^{\lfloor x \rfloor }_{k=0}(-1)^k{n\choose k}(y-k)^{n-1}dy. $$ I then tried to do this integration and came up with $$ f_{n+1}(x)=\frac{1}{n!}\left( \left[ \sum^{\lfloor x-1 \rfloor }_{k=0}(-1)^k{n\choose k}(y-k)^{n} \right]^{\lfloor x \rfloor}_{x-1} + \left[ \sum^{\lfloor x \rfloor }_{k=0}(-1)^k{n\choose k}(y-k)^{n} \right]_{\lfloor x \rfloor}^{x} \right). $$ And from this point I didn't know where to go, and I couldn't see how the binomial coefficients would combine to give the inductive step I was searching for. Am I on the right tracks at all?
