Is $ f_n=\frac{(x+1)^n-(x^n+1)}{x}$ irreducible over $\mathbf{Z}$ for arbitrary $n$?

For even $n$ we bound the function from below to show it has no real roots and thus no rational roots.

First, note that for $x\ge 0$ $$f_n(x) = \sum_{k=0}^{n-2}{n\choose k+1} x^k = n + \ldots \ge n,$$ where the function expressed by $\ldots$ is clearly positive semidefinite. (We take the sum above to be the definition of $f_n$ as implied by the problem statement, i.e, $f_n(0)=n$.) Thus, the function has no real roots for $x\ge 0$.

One can show that $$f_n(x) = \sum_{j=0}^{(n-2)/2} \frac{n(n-j-2)!}{(j+1)!(n-2j-2)!} (-x)^j (x+1)^{n-2j-2}.$$ For even $n$ and $x\in (-\infty,-1)\cup(-1,0)$ each term in the sum is explicitly positive definite. In addition, $f_n(-1)=2$. Therefore, for even $n$, $f_n$ has no real roots.