Let $L$ be the splitting field of the polynomial $ f$ over $K$, Prove that if $n!=[L:K]$, the polynomial is irreducible

Question

Let $K$ be a field and $f \in K[x]$ be a non zero polynomial of degree $n$. Let $L$ be the splitting field of $f$ over $K$.

Prove that $[L:K]$ divides $n!$ - I already proved this.

Now I am stuck at this:

Prove that if $n!=[L:K]$, the polynomial is irreducible.

I have no idea how to proceed, although I think I should go by induction, not sure how to actually carry it out.

Answer 1

Suppose not, $f=gh$ with $\deg(g)=a$ and $\deg(h)=b=n-a$ where $a,b\ge1$.

Now let $K_1$ be the splitting field of $g$ over $K$. So $[K_1:K] \le a!$ and $[L:K_1]\le b!$. So $[L:K]\le a!b!$ which is strictly less than $n!$ if any of $a,b>1$.

Edit 1: $[K_1:K] \mid a! \implies [K_1:K]\le a!$ and ${n\choose k}>1 \iff k\ne0,n$.