A random variable $X$ assumes each of the values $1,2,\ldots,m$ with probability $1/m$. Let $S_n$ be the result of sampling $n$ values of $X$ independently and summing them. Show that for $n=1,2,\ldots$, $$\mathbb P(S_n\leqslant j) = \frac 1{m^n}\sum_r (-1)^r\binom n r\binom{j-mr}n. $$
In essence, we need to find the $j$th term coefficient from the pgf of $S_n$, the sum of n observations of $X \sim U[1,m]$. I get the pgf $$G_{S_n}(t) = \frac {t^n(1-t^m)^n} {m^n} (1-t)^{-(n+1)}$$, which the answer agrees with. To get the coefficient, I put the corresponding expansions in: $$G_{S_n}(t)=\frac {t^n} {m^n} \left(\sum_{r=0}^n \binom n r (-t^m)^r \right) \left(\sum^\infty_{r=0} \frac {(-n-1)(-n-2)\cdots (-n-r)} {r!} (-t)^r \right)$$ $$=\frac {t^n} {m^n} \left(\sum_{r=0}^n \binom n r (-t^m)^r \right) \left(\sum^\infty_{r=0} \frac {(n+1)(n+2)\cdots (n+r)} {r!} t^r \right)$$ $$=\frac {t^n} {m^n} \left(\sum_{r=0}^n \binom n r (-t^m)^r \right) \left(\sum^\infty_{r=0} \binom {n+r} {n} t^r \right)$$
Now I need the Cauchy product between a finite expansion and an infinite power series but I'm not sure how to actually do the computation according to the description of it on Wikipedia because of the $t^m$ in the first sum.
Here's my attempt based off this post (probably wrong):
$$G_{S_n}(t)=\frac {t^n} {m^n} \sum^\infty_{r=0} \sum^{\min (r,k)}_{k=0} \binom n r (-t^m)^r \binom {n+r-k} {n}t^{r-k}$$
but it looks nothing like the answer so I don't think I have done it right.
Managed to figure out how to do the Cauchy product in the end.
$$G_{S_n}(t)=\frac {t^n} {m^n} (1-t^m)^n (1-t)^{-n-1}$$ $$=\frac {t^n} {m^n} \left(\sum_{r=0}^n \binom {n} {r} (-1)^r t^{mr} \right) \left(\sum^\infty_{k=0} \frac {(-n-1)(-n-2)\cdots(-n-k)} {k!} (-t)^k\right)$$ $$=\frac {t^n} {m^n} \left(\sum_{r=0}^n \binom {n} {r} (-1)^r t^{mr} \right) \left(\sum^\infty_{k=0} \binom {n+k} {n} t^k\right)$$
Then do the Cauchy product: $$=\frac {t^n} {m^n} \sum_{\alpha=0}^\infty t^\alpha \sum_{mr+k=\alpha} (-1)^r \binom {n} {r} \binom {n+k} {n}$$
The question asks for the $j$th coefficient so using Wilf's notation of $[t^j]$ to indicate the $j$th coefficient of the series:
$$[t^j]G_{S_n}(t)=\frac {1} {m^n} [t^{j-n}] \sum_{\alpha=0}^\infty t^\alpha \sum_{mr+k=\alpha} (-1)^r \binom {n} {r} \binom {n+k} {n}$$ $$=\frac {1} {m^n} \sum_{mr+k=j-n} (-1)^r \binom {n} {r} \binom {n+k} {n}$$
$mr+k=j-n$ rearranges to $n+k=j-mr$ and the summand is in terms of $r$ so $r$ is used for the limits:
$$\mathbb{P}[S_n \leq j] = [t^j]G_{S_n}(t)=\frac {1} {m^n} \sum_{r \geq 0} (-1)^r \binom {n} {r} \binom {j-mr} {n}\quad\blacksquare$$
Comtet's "Advanced Combinatorics" (1974) helped with its description of polynomial coefficients but this was easier than I made it out to be.