$5|\#\text{Gal}(f/\mathbb{Q})\subset S_5 \implies \text{Gal}(f/\mathbb{Q})$ contains a $5$-cycle?

Context: Consider $$f(x):=x^5-4x+2\in\mathbb{Q}[x].$$ By Eisenstein's criterion, $f$ is irreducible over $\mathbb{Q}$. Since $\mathbb{Q}$ has characteristic $0$, we know every irreducible polynomial in $\mathbb{Q}[x]$ is separable. In particular, $f$ is separable. We also know that the splitting field of a separable polynomial over a field $F$ is a Galois extension of $F$. Hence, if we denote by $E$ the splitting field of $f$, we have that $E$ is a Galois extension of $\mathbb{Q}$.

Now let $\alpha$ be any root of $f$. Then we have the tower of fields $$E/\mathbb{Q}(\alpha)/\mathbb{Q}.$$ Since $f$ is an irreducible and monic polynomial such that $f(\alpha)=0$, $f$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$, and we know $[\mathbb{Q}(\alpha):\mathbb{Q}]=\deg(f)=5$ (in fact even if $f$ was not monic this would be true...). Also, we know we must have $$[E:\mathbb{Q}]=[E:\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}].$$ Therefore $5|\#\text{Gal}(f/\mathbb{Q})$. It is known that $\text{Gal}(f/\mathbb{Q})\subset S_{\deg(f)} = S_5$, where $\subset$ really means that there exists a monomorphism from $\text{Gal}(f/\mathbb{Q})$ to $S_5$.

Question: Why is it that $\text{Gal}(f/\mathbb{Q})$ (when identified to a subgroup of $S_5$) must contain a $5$-cycle? It looks like this should be a consequence of the fact that $5|\#\text{Gal}(f/\mathbb{Q})$, but I can't see why.

Note that the $5$-cycles are the only elements of $S_5$ which have order $5$ and as $5$ divides $\#\text{Gal}(f/\mathbb{Q})$, your galois group has to contain an element of order $5$ by Cauchy's theorem, hence has to contain a $5$-cycle.