On the proof of $\;(1+x)^p\equiv1+x^p \pmod p$ I know the proof for $(1+x)^p\equiv 1+x^p\mod p$ using the binomial theorem. Moreover, I know that $x^p \equiv x \mod p$ due to Fermat's theorem.
Hence, is $(1+x)^p\equiv(1+x)\equiv1+x^p \mod p$ a correct proof of this relation?
After thinking about it for a bit, one of the proof of Fermat's theorem uses the binomial theorem so my comments might have been redundant (although Fermat's theorem can be deduced from Lagrange's theorem).
I guess you can try to prove Fermat's by Pigeon hole:
Assuming $a\not\equiv 0$
$$a^i,\;\;1\le i \le p$$ takes $p$ values but $\mod p$ can only take $p-1$ distinct values ($a^p \not \equiv 0$ unless $a=0$) and so $a^i\equiv a^j$ and take inverses and the rest follows. If $a\equiv 0$, then the statement follows.
This can be extended to $(1+x)^{p^n}\equiv (1+x^{p^n})$ without using induction since $p |p^n$, then $$(1+x)^{p^n}\equiv (1+x)^{p\cdot{p^{n-1}}}\equiv 1+x \equiv 1+x^{p^n}$$
I have a gut feeling that says I overlooked something important.
If this is correct, what additional instructive value does the proof using the binomial theorem has that the Fermat's proof doesn't?
PS: I don't know what to say on the title. Feel free to edit it.
 A: $$(1+x)^p = 1+x^p$$
is true not only for $x\in \Bbb{F}_p$ but also for $x\in \Bbb{F}_p[t]$ or any commutative ring of charactistic $p$.
In contrary $a^p = a$ is true only for $a\in \Bbb{F}_p$ (as $t\ne t^p$ in $\Bbb{F}_p[t]$)
A: That's not only valid (albeit only for $\Bbb F_p$) but the neatest proof of it I've ever seen, much more cunning than using the binomial theorem (which, as @reuns & @JyrkiLahtonen noted, covers $\Bbb F_p(t)$ on a commutative ring of characteristic $p$).
A: I would read the congruence
$$(1+x)^p\equiv1+x^p\pmod p$$
to involve polynomials in the ring $\Bbb{Z}[x]$.
Two polynomials $a(x)=a_0+a_1x+\cdots+a_nx^n$ and $b(x)=b_0+b_1x+\cdots+b_mx^m$ with integer coefficients are said to be congruent modulo $p$, if and only if $a_i\equiv b_i$ for all $i$ in the range $[0,\max\{m,n\}]$.
You must not think of $x$ as an integer. Instead, it is an indeterminate. Meaning that $x^p$ and $x$ are not congruent modulo $p$. For example their linear terms have coefficients $0$ and $1$ respectively. And $0\not\equiv1\pmod p$.
Observe that, by the binomial theorem,
$$
(1+x)^p\equiv 1+x^p\pmod p
$$
even though $x$ and $x^p$ are not congruent modulo $p$.

Another way to look at this is to say that by the binomial theorem we have the identity (not a congruence!) in the polynomial ring $\Bbb{Z}_p[x]$:
$$
(1+x)^p=1+x^p.
$$
Observe that, for example, in the polynomial ring $\Bbb{Z}_2[x]$ we don't have the identity $x=x^2$. The polynomial on the left has degree one whereas the polynomial on the right hand side has degree two.

Yet another way to see the difference is to let $x$ take values from e.g. the ring of Gaussian integers $\Bbb{Z}[i]$. For example, with $p=3, x=1+i$ we see that
$$(1+i)^3=-2+2i.$$
This is not congruent to $(1+i)$ modulo $3$, so $x\not\equiv x^3\pmod 3$ for this choice of $x$. But, we have
$$
(1+x)^3=(2+i)^3=2+11i
$$
and
$$
1+x^3=1+(1+i)^3=-1+2i.
$$
Their difference is
$$
(1+x)^3-(1+x^3)=3+9i=3(1+3i)
$$
a multiple of three. All in line with the congruence
$$(1+x)^3\equiv1+x^3\pmod3.$$
