 Solution

The multiplicative group $K^{\times }$ has order $p^n-1$, so Lagrange's theorem for finite groups gives $x^{p^n-1}=1$ for $x\ne 0$. Including zero, $x^{p^n}=x$ for every $x\in K$. Therefore ${\vartheta }^n$ is the identity. No positive smaller power is the identity by part (iii), soFor $n=1$ this says that Frobenius itself is the identity, as it should be.