The polynomial f(X) has integer coefficients and an integer root. Prove that, for every non negative integer n, the product f(0)f(1)...f(n) is divisible by (n+1)! .
Just some ideas ...
The integer root is very important. You can represent a polynomial as $\displaystyle f(x) = \sum_{i=0}^n a_i x^i$ where the $\displaystyle a_i$ are integers (as stated in the problem). How would you represent the polynomial if you knew one of its roots, say $\displaystyle r$?
What would happen to the product $\displaystyle f(0)f(1) ... f(n)$ if $\displaystyle r \geq 0$ (i.e. $\displaystyle r$ is also nonnegative) and $\displaystyle n \geq r$? what about if $\displaystyle r \geq 0$ and $\displaystyle n < r$? Remember, $\displaystyle r$ is an integer root of the polynomial, and $\displaystyle n$ is nonnegative.
What about similar situations for when $\displaystyle r < 0$ (i.e. what are the possible cases if $\displaystyle r$ is negative?
That should help you get started...