The probability distribution function of uniform random variables is as given Given $U_1, U_2, \dots, U_n$ where each $U_i \sim U[0,1]$, then use uniqueness theorem to show probability distribution function of $X = U_1 + U_2 + \ldots +U_n$ (sum of independent uniform random variables) is:
$$f(x) = \frac{1}{(n-1)!}\sum_{k=0}^n(-1)^k\binom{n}{k}(x-k)^{n-1}_{+}$$
The uniqueness theorem says the characteristic function uniquely determines the probability distribution. The characteristic function for $U_i$ in this case will be $\phi_i(x)=$ $\int_{0}^{1} e^{itx}dx = \frac{e^{it}-1}{it}$. Next lets find the characteristic function for X, this will be:
$$\phi_X(t)$ = E(e^{it(U_1+\ldots+U_n)}) = \left(\frac{e^{it}-1}{it} \right)^n $$
From discussion in comments, to apply uniqueness theorem here, we consider a random variable Y such that Y has distribution f(x). Then $\phi_Y(t) = E(e^{iYt}) = \int_{0}^nf(x)e^{ixt}dx = \sum_{k=0}^{n} (-1)^k\binom{n}{k}\frac{1}{(n-1)!}\int_{0}^{n}(x-k)^{n-1}e^{ixt}dx$. I am not sure how the computation goes from here, but we want to show $\phi_Y(t) = \phi_X(t)$.
 A: Note, that your last expression
$$\phi_Y(t) = \sum_{k=0}^{n} (-1)^k\binom{n}{k}\frac{1}{(n-1)!}\int_{0}^{n}(x-k)^{n-1}e^{ixt}dx$$
is not exactly correct. In the sum you have $(x-k)_+$ which is $0$ unless $x>k$. So it should be
$$\phi_Y(t) = \sum_{k=0}^{n} (-1)^k\binom{n}{k}\frac{1}{(n-1)!}\int_{k}^{n}(x-k)^{n-1}e^{ixt}dx$$
Now, let $u=x-k$ and $J_{n,m}=\int_{0}^{n-k}u^{n-m}e^{iut}du$:
$$\phi_Y(t) = \sum_{k=0}^{n} (-1)^k\binom{n}{k}\frac{e^{ikt}}{(n-1)!}J_{n,1}$$
$$J_{n,1}=\frac{1}{it}\int_{0}^{n-k}u^{n-1}de^{iut}=\frac{1}{it}\left[e^{iut}u^{n-1}|_{u=0}^{n-k}-(n-1)J_{n,2}\right]=...=$$
$$=e^{iut}\sum_{j=1}^n(-1)^{j-1}\frac{(n-1)!}{(n-j)!(it)^j}u^{n-j}|_{u=0}^{n-k}=$$
$$=e^{i(n-k)t}\sum_{j=1}^n(-1)^{j-1}\frac{(n-1)!}{(n-j)!(it)^j}(n-k)^{n-j}+(-1)^n\frac{(n-1)!}{(it)^n}$$
Now plug this into $\phi_Y(t)$, change summation order in the first term and note that $\sum_{k=0}^n(-1)^k\binom{n}{k}(n-k)^m=0$ for $m<n$. So that the first term is just $0$. What is left is just:
$$\phi_Y(t) = \sum_{k=0}^{n} (-1)^k\binom{n}{k}\frac{e^{ikt}}{(n-1)!}(-1)^n\frac{(n-1)!}{(it)^n}=\frac{(e^{it}-1)^n}{(it)^n}$$
