I have no idea how to approach this.
I'm supposed to use $f\in \mathbb{Q}[X]$ is irreducible $\iff \exists{a}\in\mathbb{Q}$ such that $f(X+a)$ is irreducible.
I tried to use $a = 1 \in \mathbb{Q}$ so $\frac{(X + 1)^p-1}{X}=(X+1)^{p-1}+\cdots+ (X+1) + 1$ but I don't see how I can get from there to being able to use any other criterion.
From where you left off:
$\displaystyle \sum_{i=0}^{p-1} (X+1)^i = \displaystyle \sum_{i=0}^{p-1} \displaystyle \sum_{k=0}^i \binom{i}{k} X^k = 1 + (1 + X) + \left(1 + \binom{2}{1}X + \binom{2}{2}X^2\right) + \left(1 + \binom{3}{1}X + \binom{3}{2}X^2 + X^3\right) + ... + \left(1 + \binom{k}{1}X + \binom{k}{2}X^2 +...+ \binom{k}{j}X^j +..+ X^k\right) + ...+ \left(1 + \binom{p-1}{1}X + \binom{p-1}{2}X^2 +...+ \binom{p-1}{j}X^j +...+ X^{p-1}\right) = p + \displaystyle \sum_{k=1}^{p-1} \binom{k}{1}X + \displaystyle \sum_{k=2}^{p-1} \binom{k}{2}X^2 + ...+ \displaystyle \sum_{k=q}^{p-1} \binom{k}{q}X^q + ...+ X^{p-1}$. You can use the Eisenstein's criterion for this polynomial.