Power to Characteristic Power of Field is Monomorphism

Theorem
Let $F$ be a field whose characteristic is $p$ where $p \ne 0$.

Let $n \in \Z_+$ be any positive integer.

Let $\phi_n: F \to F$ be the mapping on $F$ defined as:
 * $\forall x \in F: \phi_n \left({x}\right) = x^{p^n}$

Then $\phi_n$ is a (field) monomorphism.

Proof
Proof by induction:

For all $n \in \Z_+$, let $P \left({n}\right)$ be the proposition:
 * $\phi_n$ is a (field) monomorphism.

$P(0)$ is trivially true:
 * $\phi_0 \left({x}\right) = x^{p^0} = x^1 = x$

... and we see that $\phi_0$ is the identity automorphism.

This is not the zero homomorphism.

So from Homomorphism from Field Either Monomorphism or Zero Homomorphism, it follows that $\phi_0$ is a ring monomorphism.

Basis for the Induction
First we need to show that $P(1)$ is true:
 * $\phi_1 \left({x}\right) = x^{p^1} = x^p$ is a (field) monomorphism.

This is demonstrated to be a monomorphism in Power to Characteristic of Finite Field is Monomorphism.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $\phi_k \left({x}\right) = x^{p^k}$ is a (field) monomorphism.

Then we need to show:
 * $\phi_{k+1} \left({x}\right) = x^{p^{k+1}}$ is a (field) monomorphism.

Induction Step
This is our induction step:

Multiplication is more straightforward:

... and does not rely on the induction process.

$\phi_{k+1}$ is not the zero homomorphism.

So from Homomorphism from Field Either Monomorphism or Zero Homomorphism, it follows that $\phi_{k+1}$ is a ring monomorphism.

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \Z_+: \phi_n$ is a (field) monomorphism.

Also see

 * Prime Power of Sum Modulo Prime, where the same technique is used.