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Using the Binomial Theorem $(x+1)^p$ can be written out as
$$
(x+1)^p = sum_k=0^p binompk x^k,.
$$
All that's left to show then is that, if $q$ is a proper prime factor of $p$, then one of $binomp1, dots, binompp-1$ isn't divisible by $q$. For that consider $binompq^t$ where $q^t$ is the biggest power of $q$ that occurs in $p$. This coefficient is
$$
binompq^t = fracp(p-1) cdots (p-q^t+1)q^t(q^t-1) cdots 1 = q bigg(q^t-1kfrac(p-1) cdots (p - q^t + 1)q^t(q^t-1) cdots 1 bigg)
$$
for some integer $k$. So, if we can just show that this big parenthesis doesn't work out to an integer, we're done. And I feel like there ought to be a quick way to finish the argument. But the only way I see to proceed is to painstakingly count the number of occurrences of $q$ in both the numerator and denominator and show that there aren't enough in the numerator to fully cancel the denominator. That's quite messy. Is there a more elegant maneuver?

In general it may actually be the case that indeed $(x+1)^pequiv_q x^p+1$. As an example, take $q=3,p=9$.

In fact we can classify exactly which $p$ do this; it turns out that $(x+1)^pequiv_q x^p+1$ for all $x$ if and only if $p=q(qk-k+1)$ for some $k$.

To see why this is true, let $g$ be a primitive generator of $mathbbF_q^*$, i.e., $g^n=1$ iff $q-1|n$. Suppose $(x+1)^pequiv_qx^p+1$. By plugging in $x=1$ and using induction, this would imply that $x^pequiv_q x$ for all $x$, so $x^p-1equiv_q 1$. In particular, $g^p-1equiv_q 1$, so by definition $q-1|p-1$.

Writing out $p=qcdot m$, this means there exists some $n$ such that $n(q-1)=mq-1$. Then $(n+(q-1))(q-1) = (m+(q-2))q$. Since $q$ cannot divide $q-1$ and $q$ is prime, this means $q|n+q-1$, so $q|n-1$. Thus we can write $n=qk+1$, which gives $mq-1 = (qk+1)(q-1)$, so expanding out the terms gives $m=qk-k+1$.

Conversely, if $p$ can be written in this form, then since every element $xinmathbbF_q^*$ has order dividing $q-1$, we get $x^p=x^q(qk-k+1)equiv_q x^kx^-kxequiv_q x$, from which the stated equality immediately holds.

EDIT. To answer your other question of whether it is true that $binompq^tnotequiv_q 0$ where $q^tmid p$ but $q^t+1nmid p$: This can be proven as follows.

Let $m,n$ be arbitrary integers, and consider $binommnm=frac(mn)(mn-1)cdotsleft(mn-(m-1)right)m!$. We can remove a factor of $m$ from the top and bottom and are left with $binommnm=fracn(mn-1)cdotsleft(mn-(m-1)right)(m-1)!$. Now if $k$ is any factor of $m$ that divides $mn-j$ for some $j$, then $k$ must also divide $j$. Thus we can remove factors from the top and bottom until we're left with $binommnm=fracnpq$ where $mnmid p$. Thus if $m$ divides $binommnm$ it must also divide $n$.

Now substituting $m=q^t$ and $mn=p$ immediately gives the desired result, since $q^tmidbinompq^timplies q^tmidfracpq^timplies q^2tmid p$, contradicting maximality of $t$.