Why is this hypothesis necessary? The following seems to be a common lemma for proving Fermat's little theorem:

Let $1\leq k\leq p-1$ and $p$ be  a prime. Then $\binom{p}{k}\equiv 0\pmod p$.

We have $$\binom{p}{k}=\frac{p!}{k!(p-k)!}.$$ It seems that $p$ being prime is completely unecessary to get $p\,| \binom{p}{k}$. This proof states:

"...since $p$ is prime, then $p$ divides the numerator, but not the denominator."

That doesn't make sense to me...
What is the purpose of having "$p$ is a prime" part of the hypotheses?
 A: Hint
We have
$$k!(p-k)!\binom{p}{k}=p!$$
Now $p$ divides $k!(p-k)!\binom{p}{k}$ but doesn't divide $k!(p-k)!$ (why?) and use the Euclid lemma.
A: $4\choose 2$ $=\frac{4!}{2!2!}= 6$ is not a multiple of 4.  Since 4 is not prime, it chops up into pieces (2x2), and some of those pieces can be taken away by the denominator.
A: The reason that we know the $p$ does not divide the numerator is because $p$ is prime. 
$k!(p-k)!$ is a product of terms which are all less than $p$, so $p$ cannot divide it. 
One way to see this would be to use one of the key features of primes, that if $p|ab$, then $p|a$ or $p|b$. If $p|k!(p-k)!$, since $p\not|k$, it must be the case that $p|(k-1)!(p-k)!$. Now we can repeat this process for each term and eventually conclude that $p|(p-k)$, a clear contradiction.   
Another way to see this would be by writing $k!(p-k)!$ in it's unique prime factorisation, which is known to be the product of the unique factorisations of each term. Clearly $p$ does not divide any individual term as they are all less than $p$, so it is not in any of the factorisations of the individual terms, so it cannot be in the factorisation of the whole thing and thus does not divide it.
The reason that these approaches both worked is because of the fact that $p$ was prime. Indeed, if $p$ is not prime then there exists $1\leq k\leq p-1$ such that $p\not|\binom{p}{k}$ (You might like to think about why). In this sense, not only is the hypothesis sufficient for the lemma, it is actually necessary in the strict mathematical sense.
(For completeness, I'll quickly note that both of the above approaches immediately generalise to prove the following theorem):

For any prime $p$, if $a_1,a_2,\dots,a_n$ is a sequence of natural numbers with $1\leq a_i\leq p-1$ for each $i$, then $p\not|\prod_{i=1}^{n}a_i$

