Let $p\in\mathbb{P}$. Consider the polynomial
$$f(t) = (t-1)(t-2)\ldots (t-(p-1)) -t^{p-1}+1 $$
Show that every coefficient is divisible by $p$.
The first thing to notice is that we have $p-1$ factors, therefore the summands $t^{p-1}$ cancel or in other words, the coefficient for $t^{p-1}$ is zero and is divisible by $p$.
Multiplying everything through seems veeery tedious. Can we get by easier, somehow?
We would have something like this:
$$f(t) = 0t^{p-1}+A_1t^{p-2} + A_2t^{p-3} +\ldots A_{p-2}t +(1+\prod_{i=1}^{p-1}i) $$
We could suppose for a contradiction that one of the coefficients is not congruent to $0$ modulo $p$, but this doesn't help, because, for one - there is no information on congruence of $f(t)$ and secondly, why would the constant term have to satisfy that congruence in general?
..or have I mis-understood the problem? How to proceed?
This is a very cute problem. Let me make one innocuous observation before I begin:
A non-zero polynomial of degree $n$ defined over a field can have at most $n$ roots.
Now let us begin: Plug in various values for $t$ in $f(t)=(t-1)(t-2)\cdots (t-(p-1))-t^{p-1}+1$. We see that $f(0)=(p-1)!+1 \equiv0 \pmod p$ by Wilson's theorem. For any other value of $t$ modulo $p$, i.e. $1\leq t_0 \leq p-1$, we notice that the product $(t_0-1)(t_0-2)\cdots (t_0-(p-1))$ is zero because $t_0-t_0$ appears in one of the terms in the product. So $f(t_0)=0-t_0^{p-1}+1 \equiv 0 \pmod p$ by Fermat's little theorem!
So the polynomial $f(t)$ has $p$ roots modulo $p$. However it's degree is $\leq p-1$. By the observation I made at the beginning, this forces $f$ to be the zero polynomial modulo $p$. This means that every coefficient of $f$ is divisible by $p$.