Problem: Let $L\mid K$ be a finite extension of fields. Then $K$ is perfect $\iff$ $L$ is perfect.
The implication $\implies$ is quite easy and it has already been discussed here.
I'm interested in the implication $\impliedby$ because of the following application:
Suppose $\text{char}(k)=p>0$. Then any finitely generated field extension $K\mid k$ that is perfect has $\text{tr.deg}_k(K)=0$.
For the proof of this recall that a purely trascendental extension is never perfect as $t_1$ is never a $p$-power in the field $k(t_1,\dots,t_n)$. Then if $K$ is a finite extension of a purely trascendental extension it wouldn't be perfect as well.
That can be restated in a neat way in the language of algebraic geometry:
If $\text{char}(k)>0$ then any $k$-variety with perfect function field must have dimension $0$.
Also it would be interesting to see if the equivalence in the problem is true if we change finite extension by algebraic extension. Again the implication $\implies$ is not so hard and has been discussed here.
Best Answer
Suppose $K$ is not perfect. Then some $a\in K$ has no $p$th root in $K$. I claim that the polynomial $f(x)=x^{p^n}-a$ is irreducible over $K$ for any $n\in\mathbb{N}$. To prove this, let $b$ be a $p^n$th root of $a$ in an extension field of $K$ and note that $f(x)$ factors as $(x-b)^{p^n}$. Let $g$ be the minimal polynomial of $b$ over $K$. Then $g$ is the minimal polynomial of every root of $f$, so $f=g^m$ for some $m$. Since $\deg f=p^n$, $m$ must be a power of $p$; say $m=p^d$. We then have $g(x)=(x-b)^{p^{n-d}}=x^{p^{n-d}}-b^{p^{n-d}}$ and thus $b^{p^{n-d}}\in K$. If $d>0$, we see that $(b^{p^{n-d}})^{p^{d-1}}=b^{p^{n-1}}$ is a $p$th root of $a$ in $K$, which is a contradiction. Thus $d=0$ and $m=1$ so $f=g$ is irreducible.
Now if $L$ is a perfect extension of $K$, then $a$ must have a $p^n$th root in $L$ so $f$ must have a root in $L$. Since $f$ is irreducible, this means $[L:K]\geq \deg f=p^n$. Since $n$ is arbitrary, this means $[L:K]$ must be infinite.