If $f(x)\in\mathbb{Q}[x]$ of degree $p$ and $\operatorname{Gal}(K/\mathbb{Q})$ has element of order $p$ then $f(x)$ is irreducible. Let $f(x)\in\mathbb{Q}[x]$ , $p$ prime, $\deg f(x)=p$ and $G = \operatorname{Gal}(K/\mathbb{Q})$ has element of order $p$, where $K$ the the splitting field of $f(x)$ over $\mathbb{Q}$.
Show that $f(x)$ is irreducible over $\mathbb{Q}$.

Here's a very rough sketch of how to get started:
Suppose for contradiction $f$ were reducible.  Then $f(x) = g(x)h(x)$, such that $\deg(g) + \deg(h) = p$.
Choose any root $\alpha_1$ of $g$.  then $\mathbb{Q}[\alpha_1]$ is a field extension of $\deg(g) <p$ degree.  Suppose that, in this new extension, $g(x)$ now splits into $g(x) = \displaystyle \prod_{i = 1}^{k}g_k(x)$.  If we continue adjoining roots of the irreducible $g_k$'s to $\mathbb{Q}$ until $g$ has been completely split, we obtain the following tower of fields:
$$\mathbb{Q} \subset \mathbb{Q}[\alpha_1] \subset \mathbb{Q}[\alpha_1, \alpha_2] \subset ... \subset \mathbb{Q}[\alpha_1, \alpha_2, ..., \alpha_n]$$
Since each successive extension has degree strictly less than $p$, we know the degree of the final extension must have a degree not divisible by $p$. (See foot-note)
If $h(x)$ has not been completely split in this new extension, then simply continue the process for it.  The end result will be the same: the final extension cannot have degree divisible by $p$.
Hence: $p \nmid [K:\mathbb{Q}]$.  Therein lies a contradiction: since $\operatorname{Gal}(K/\mathbb{Q})$ has an element of order $p$, then we know $p$ divides $|\operatorname{Gal}(K/\mathbb{Q})|$.  However, it is a theorem that $|\operatorname{Gal}(K/\mathbb{Q})| = [K:\mathbb{Q}]$.



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*This is coming from the fact that, given a tower of fields $F \subset E \subset K$, then $[K:F] = [K:E]\cdot[E:F]$.

