Show that the polynomial $(x-1)(x-2) \cdots (x-n)-1$ is irreducible on $\mathbb{Z}[x]$ for all $n \geq 1$ 
Show that the polynomial $h(x)=(x-1)(x-2) \cdots (x-n)-1$ is irreducible in $\mathbb{Z}[x]$ for all $n \geq 1$.

This problem seems to be hard to solve. I thought I could use Eisenstein in developping this polynomial, but it is a bad idea. Another idea would be to suppose the existence of $f(x)$ and $g(x)$, and suppose that $f(x)g(x)=(x-1)(x-2) \cdots (x-n)-1$. In this direction, we could analyse the roots of $h(x)$ I guess.
Is anyone could help me to solve this problem?

David's observation that if $f=gh$, then $g(k)=-h(k)=1$ or both $=-1$ for each of $k=1,2,\ldots , n$ is spot on. So both $g$ and $h$ take at least one of these values at least $n/2$ times. If we now take the polynomial of smaller degree (let's say it's $g$), so $\deg g=m\le n/2$, then the only way to avoid a constant $g$ would be $m=n/2$ (so we're done already if $n$ is odd) and a polynomial of the type
$$
g = (x-k_1)(x-k_2) \ldots (x-k_m)+ 1 .
$$
But we also have to have $g(j)=-1$ for $1\le j\le n$, $j\not= k_r$, and this clearly isn't working.
