Is $ f_n=\frac{(x+1)^n-(x^n+1)}{x}$ irreducible over $\mathbf{Z}$ for arbitrary $n$? 
In this document on page $3$ I found an interesting polynomial:
  $$f_n=\frac{(x+1)^n-(x^n+1)}{x}.$$
  Question is whether this polynomial is irreducible over $\mathbf{Q}$ for arbitrary $n \geq 1$.

In the document you may find a proof that polynomial is irreducible whenever $n=2p , p \in \mathbf{P}$ and below is my attempt to prove that polynomial isn't irreducible whenever $n$ is odd number greater than $3$. Assuming that my proof is correct my question is :
Is this polynomial irreducible over $\mathbf{Q}$ when $n=2k , k\in \mathbf{Z^+}$ \ $\mathbf{P} $ ?

Lemma 1: For odd $n$, $a^n + b^n = (a+b)(a^{n-1} - a^{n-2} b + \cdots + b^{n-1})$.
Binomial expansion: $(a+b)^n = \binom{n}{0} a^n + \binom{n}{1} a^{n-1} b + \cdots + \binom{n}{n} b^n$.
$$\begin{align*} f(x)  &= \frac{(x+1)^n - x^n-1}{x} = \frac{(x+1)^n - (x^n+1)}{x}
\\ &= \frac{(x+1) \cdot (x+1)^{n-1} - (x+1)(x^{n-1}-x^{n-2}+ \cdots + 1)}{x} 
\\ &= \frac{(x+1) \cdot (x+1)^{n-1} - (x+1)(x^{n-1}-x^{n-2}+ \cdots + 1)}{x} 
\\ &= \frac{(x+1) \cdot \left[ \Bigl(\binom{n-1}{0}x^{n-1}+ \binom{n-1}{1} x^{n-2} + \cdots + 1 \Bigr) - (x^{n-1}-x^{n-2}+ \cdots + 1) \right]}{x} 
\\ &= \frac{(x+1) \cdot \left[ \Bigl(\binom{n-1}{0}x^{n-1}+ \binom{n-1}{1} x^{n-2} + \cdots + \binom{n-1}{n-2} x \Bigr) - (x^{n-1}-x^{n-2}+ \cdots - x) \right]}{x} 
\\ &= (x+1) \cdot \small{\left[ \left(\binom{n-1}{0}x^{n-2}+ \binom{n-1}{1} x^{n-3} + \cdots + \binom{n-1}{n-2}  \right) - (x^{n-2}-x^{n-3}+ \cdots - 1) \right]} \end{align*}$$
So, when $n$ is an odd number greater than $1$, $f_n$ has factor $x+1$. Therefore, $f_n$ isn't irreducible whenever $n$ is an odd number greater than $3$. 

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.
