TITLE: If $f(x)$ is an irreducible polynomial of degree n, then the cardinality of its Galois group is divisible by $n$.
QUESTION [6 upvotes]: 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.

REPLY [6 votes]: 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.