It seems that all coefficients, except the first and the last one, of the polynomial $(x-1)(x-2)...(x-(p-1))$, where $p$ is a prime, are divisible by $p$. Example for $p=7$:
$(x-1)(x-2)(x-3)(x-4)(x-5)(x-6) = x^6 - 21 x^5 + 175 x^4 - 735 x^3 + 1624 x^2 - 1764 x + 720$
The coefficients 21, 175, 735, 1624, 1764 are all divisible by 7.
How to prove this for all $p$?
The coefficient of $x^{p-2}$ is $-(1 + 2 + ... + (p-1)) = -p(p-1)/2$ which is divisible by $p$ for $p>2$. (For $p=2$ the corresponding polynomial is $x-1$, which does not have any coefficients except the first and last one.) Is there a general formula for the coefficients of this kind of polynomial with roots 1, ..., $(p-1)$ (from which the divisibility follows)?
These coefficients are by definition the Stirling numbers of the first kind (see also Coefficients of $(x-1)(x-2)\cdots(x-k)$), as noted in comments. Now notice that $$x^{p} - x \equiv x(x-1)(x-2)(x-3)\cdots (x -(p-1))\pmod{p},$$ since left hand side is divisible by $p$ for all integers $x$ by Fermat's little theorem, as is the right hand side (it is a product of $p$ consecutive integers). And so $$ [x^i]P(x)=[x^{i+1}](xP(x))\equiv[x^{i+1}](x^p-x)\pmod p,$$ and the result follows.