All of the coefficients of $P(x)$ are divisible by $p$ 
Find a counterexample to the following: If $P(x)$ is a polynomial with integer coefficients and $p$ a prime, then if $p \mid P(x)$ for all $x$ then all of the coefficients of $P(x)$ are divisible by $p$.

I thought about proving this by a proof by contradiction but wasn't sure how to do that. How should we prove the statement?

Another counterexample: any sequence of $p$ consecutive integers contains a multiple of $p$, so $P(x)=x(x+1)(x+2)...(x+p-1)$ is always a multiple of $p$ for integer $x$, while the leading coefficient is $1$ which is obviously not divisible by $p$.
