If $f(x)$ is an irreducible polynomial of degree n, then the cardinality of its Galois group is divisible by $n$. 
If $f(x)$ is an irreducible polynomial of degree $n$, then the cardinality of its Galois group is divisible by $n$.

I know I need to use the Tower Theorem, but I can't figure out how to get from there to the solution.
 A: Here is another solution (the one by thyde641 is just fine, of course): 
It is easy to show that the Galois group of $f$ acts transitively on the roots of $f$  (here, irreducibility is crucial). We moreover suppose $f$ to be separable, so that it has $n$ distinct roots in its splitting field. A finite group which acts transitively on a set of $n$ elements has order divisible by $n$, by the orbit-stabilizer theorem. 
A: Let $\alpha$ be a root of $f(x)\in K[x]$ in an algebraic closure $\overline{K}$. Let $L$ be the splitting field of $f(x)$. Then $K \subseteq K(\alpha) \subseteq L$, since $L$ contains all the roots of $f(x)$. The size of the Galois group of $f(x)$ is equal to $[L:K]$ (assuming $f(x)$ separable, which is typical.) Now we apply the tower theorem:
$$K \subseteq K(\alpha) \subseteq L \Longrightarrow [L:K] = [L:K(\alpha)][K(\alpha):K].$$
Since $f(x)$ is irreducible of degree $n$, it follows that $[K(\alpha):K]=n$. Stringing all this together, we have
$$\#\operatorname{Gal}(L/K) = [L:K] = [L:K(\alpha)][K(\alpha):K] = [L:K(\alpha)]n,$$
which is the desired conclusion.
