Show that $a^{p^n}=a\mod p$ 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 $X^{p^n}-X$.
 A: 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 $\,f^{\large n}(x) = x^{\large p^{\Large n}}\pmod {p}$
A: 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
$\alpha^{p^n} = \alpha, \; \forall n \in \Bbb N. \tag{12}$
