It's well-known that $f$ has infinitely many prime divisors. Let $p_{i,j},1\le i\le n$ and $1\le j\le k$ be a sequence of distinct primes, each much larger than $n$, with $u_{i,j}$ such that $f(u_{i,j})\equiv 0\pmod{p_{i,j}}$. Next, take $a\equiv -i+u_{i,j}\pmod{p_{i,j}}$, where such $a$ exists by the Chinese Remainder Theorem. Consider now the numbers $a+1,a+2,\dots,a+n$. We have $f(a+i)\equiv f(u_{i,j})\equiv 0\pmod{p_{i,j}}$, so $f(a+i)$ is divisible by $p_{i,1},\dots,p_{i,k}$, for every $i$, completing the proof.