Is $(a+b)^{p^n} \equiv a^{p^n}+b^{p^n} \pmod{p}$ true? Is $(a+b)^{p^n} \equiv a^{p^n}+b^{p^n} \pmod{p}$ true?
Here $p$ is a prime number. 
If so, how to prove it?
I know the statement is true when $n=1$. But I have no idea about the case when $n>1$. 
The question is derived from the statement:

The solutions of $x^{p^n}=x$ forms a subfield of the field $\Bbb{F}_{p}$.

 A: You know for $k\leq p$, $p\mid \binom{p^n}{k}$
Now for $k>p$
$$\binom{p^n}{k}=\binom{p^n-1}{k}+\binom{p^n-1}{k-1}=\binom{p^n-2}{k}+2\binom{p^n-2}{k-1}+\binom{p^n-2}{k-2}\\=\dots =\binom{p^n-p}{k}+\binom{p}{1}\binom{p^n-p}{k-1}+\binom{p}{2}\binom{p^n-p}{k-2}+\dots +\binom{p^n-p}{k-p}$$
$p\mid \binom{p}{j}$, $i\leq j\leq p-1$. Continue the process on $\binom{p^n-p}{k}$.
A: Overkill, but the Frobenius map $x \mapsto x^p$ is an endomorphism of $\mathbb{F}_p$ - this is the statement for $n=1$. Thus its self-composition $n$ times is also an endomorphism.

Another overkill: by Fermat's little theorem,
$$c^{p^n} \equiv (c^p)^{p^{n-1}} \equiv (c)^{p^{n-1}} \equiv \dots \equiv c \pmod p$$
for all $c \in \mathbb{F}_p$, hence $(a+b)^{p^n} = a+b = a^{p^n} + b^{p^n} \pmod p$. This is just using the fact that Frob is actually the identity on $\mathbb{F}_p$.
A: Since you know the statement for $n=1$, simple induction suffices for the general case:
$$
  (a+b)^{p^{n+1}}=((a+b)^p)^{p^n} \overset{{\rm case}~n=1}\equiv(a^p+b^p)^{p^n}
  \overset{\rm I.H.}\equiv(a^p)^{p^n}+(b^p)^{p^n}=a^{p^{n+1}}+b^{p^{n+1}}.
$$
