Show that $a^{p^n}=a\mod p$

Question

My book says that for elements $\alpha$ in $\mathbb F_p$, where $p$ is prime, it holds that $$ \alpha^{p^n}=\alpha, $$ because of Fermat's little theorem, which says that $$ a^p=a\mod p. $$ Of course it's clear that $\alpha^p=\alpha$, but I don't see why this would hold for arbitrary powers of $p$. We have $$ \alpha^{p^n}=\alpha^n, $$ but this isn't necessarily equal to $\alpha$, is it?

For context: my book uses this argument, to show that the roots of $X^{p^n}-X$ in $\overline{F_p}$ form a field of $p^n$ elements, and they argue that $F_p$ is a subset of the set of roots of $X^{p^n}-X$.

Answer 1

Fixed points stay fixed under iteration, by an obvious induction:

if $\,f(x) = x\,$ then $\, \color{#c00}{f^{\large n}(x) = x}\,\Rightarrow\, f^{\large n+1}(x) = f(\color{#c00}{f^{\large n}(x)}) = f(\color{#c00}x) = x$

OP is the special case $\, f(x) := x^{\large p}\,$ so $\,f^{\large n}(x) = x^{\large p^{\Large n}}\pmod {p}$

Answer 2

Observe that, in any ring $R$, for $\alpha \in R$ and $m \in \Bbb N$ we have

$\alpha^{m^2} = (\alpha^m)^m; \tag 1$

$\alpha^{m^3} = (\alpha^{m^2})^m; \tag 2$

and in general, for $n \in \Bbb N$,

$\alpha^{m^n} = \alpha^{m^{n - 1}m} = (\alpha^{m^{n - 1}})^m. \tag 3$

Now suppose that $R$ is such that

$\alpha^m = \alpha; \tag 4$

then from (1) and (2),

$\alpha^{m^2} = (\alpha^m)^m = \alpha^m = \alpha, \tag 5$

$\alpha^{m^3} = (\alpha^{m^2})^m = \alpha^m = \alpha; \tag 6$

thus if

$\alpha^{m^j} = \alpha, \tag 7$

then

$\alpha^{m^{j + 1}} = \alpha^{m^jm} = (\alpha^{m^j})^m = \alpha^m = \alpha, \tag 7$

and by induction it may be concluded that

$\alpha^{m^n} = \alpha \tag 8$

holds for all $n \in \Bbb N$.

Now taking

$m = p \in \Bbb P, \tag 9$

and

$R = \Bbb F_p \tag{10}$

we have

$\alpha^p = \alpha; \tag{11}$

thus it follows that

$\alpha^{p^n} = \alpha, \; \forall n \in \Bbb N. \tag{12}$