Comparing coefficients in finite field We start with the wrong proof of the following theorem: $p| \binom{p}{k}$ for a prime $p$ and $0<k<p.$
Proof: $(1+x)^p \equiv 1+x \equiv 1+x^p \pmod{p}$ by Fermat's little theorem. Comparing coefficient of $x^k$ gives the result.
Theorem: $\binom{2p}{p} \equiv 2 \pmod{p}.$
Proof: $(1+x)^{2p} \equiv (1+x^p)^2 \equiv x^{2p} + 2x^p + 1 \pmod{p}$. Comparing the coefficient of $x^p$ gives the result.
I knew that the proof must be wrong because otherwise we could use the same logic (by comparing coefficient of $x$ in $1+x \equiv 1+x^p$) to deduce that $1\equiv 0$. But why can't we do so as in $\mathbb{R}[x]$? 
Also, I want to ask about the correct proof of the second theorem, thank you so much.
 A: Fermat's little theorem applies to residues of integers, not the indeterminate in a polynomial ring of positive characteristic, so you can't do $(1+x)^p=1+x$ or $x=x^p$ as polynomials in $\Bbb F_p[x]$ by little Fermat as it doesn't allow that. In fact if we extend the field $\Bbb F_p$ to $\Bbb F_q$ for some power $q$ of the prime $p$, the map $x\mapsto x^p$ is no longer the identity map like it is in $\Bbb F_p$, it is a nontrivial field automorphism (and it generates the Galois group). In a commutative ring of characteristic $p$ the map $x\mapsto x^p$ is in general a ring endomorphism but not necessarily an automorphism which is called the Frobenius map.
In fact one shows $(a+b)^p=a^p+b^p$ in characteristic $p$ by using the fact that $p\mid\binom{p}{k}$ for the values of $0< k<p$ (this is called variously "freshman's dream," "fool's binomial theorem," etc).
We show $p\mid\binom{p}{k}$ for $0<k<p$ by invoking $\binom{a}{b}=\frac{a!}{b!(a-b)!}~\Rightarrow~\binom{p}{k}=\frac{\color{Red}{p}(p-1)\cdots(p-k+1)}{k(k-1)\cdots3\cdot2\cdot1}$ and then this proves $(a+b)^p=a^p+b^p$ in any ring of characteristic $p$ which means we can expand your polynomial as $(1+x)^{2p}=(1+x^p)^2=1+2x^p+x^{2p}$ which gives $\binom{2p}{p}\equiv2$ mod $p$.
Another way of proving $p\mid\binom{p}{k}$ is to let the cyclic group $C_p$ acting on itself induce an action on the collection of $k$-subsets of itself and observe this has no fixed points if $0<k<p$ and hence all orbits have size $p$ (orbit-stabilizer) and hence the cardinality $\binom{p}{k}$ is a multiple of $p$.
Similar reasoning (applied inductively) proves that
$$\binom{n_kp^k+\cdots+n_1p+n_0}{m_kp^k+\cdots+m_1p+m_0}\equiv\binom{n_k}{m_k}\cdots\binom{n_1}{m_1}\binom{n_0}{m_0} \mod{p} $$
where $0\le n_0,\cdots,n_k,m_0,\cdots,m_k<p$. This is called Lucas' theorem and it helps us to very easily compute $\binom{n}{m}$ by writing $n$ and $m$ in base $p$ (hence the base $p$ expansions above).
