$\mathbb{F}_{16}$ does not contain a subfield of order 8

Question

Prove $\mathbb{F}_{16}$ does not contain a subfield of order 8

Suppose $\mathbb{F}_{16}$ has a subfield $F$ of order 8. I want to use the tower law to arrive at a contradiction. That is, $4=|\mathbb{F}_{16}:\mathbb{Z}_2|=|\mathbb{F}_{16}:F||F:\mathbb{Z}_2|$. I don't know how to continue.

Answer 1

We have $[F:\mathbb{F}_2]=[\mathbb{F}_{2^3}:\mathbb{F}_2]=3$, which does not divide $4$. Hence it is impossible.

Answer 2

The multiplicative group of $\mathbb{F}_{16}$ has order $15$ while that of $\mathbb{F}_8$ has order $7$. Now you can use the Lagrange's theorem.