If $\alpha$ is a root of $f$, then the minimal polynomial $p_\alpha$ of $\alpha$ divides $f$. Recall that $p_\alpha$ has degree $n=[\mathbb Q(\alpha):\mathbb Q]$, since $\mathbb Q(\alpha)\simeq \mathbb Q[x]/(p_\alpha)$.
Since $f$ is monic and a multiple of $p_\alpha$ of the same degree, it follows that $f=p_\alpha$ so that $f$ is irreducible. However, notice that we've used that $f$ is the minimal polynomial to deduce irreducibility -- I'm not sure about proving irreducibility first, without using some variant of the fact that $f$ must be the minimal polynomial.
Edit: Alternately, we can note that if $f(x)=g(x)h(x)$ for $g$, $h$ nonconstant polynomials, then $\alpha$ is a root of either $g$ or $h$. This implies that the minimal polynomial of $\alpha$ over $\mathbb Q$ has degree less than $[\mathbb Q(\alpha):\mathbb Q]$, a contradiction.