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 $3$.