<s>[INST] Prove that If $n,k\in\mathbb{N}$ and $\binom{n}{k}$ is a prime number, then $k=1$ or $k=n-1$. My attempt was the following:
Suppose $p=\binom{n}{k}$ is prime. By definition,
\begin{align}
            p&=\binom{n}{k}\nonumber\\
            p&=\frac{n}{k}\binom{n-1}{k-1}\nonumber\\
        \end{align}
Since $p,\binom{n-1}{k-1}\in\mathbb{Z}$, it follows that $k\mid n$ or $k\mid\binom{n-1}{k-1}$. We will look into both cases separately.
Case 1. Suppose $k\mid n$. Thus, $n=kx$ for some integer $x$. Hence, $p=\frac{kx}{k}\binom{n-1}{k-1}=x\binom{n-1}{k-1}$. Since $p$ is prime, it follows that $x=1$ and $\binom{n-1}{k-1}=p$ or $x=p$ and $\binom{n-1}{k-1}=1$. If $\binom{n-1}{k-1}=p=\binom{n}{k}$, it follows that $n=k$ and $p=1$ which is not prime, so it must be the case that $x=p$ and $\binom{n-1}{k-1}=1$, which implies $k=1$ or $k=n-1$.
Case 2. Suppose $k\mid\binom{n-1}{k-1}$. Thus, $\binom{n-1}{k-1}=ky$, for some integer $y$. Hence, $p=\frac{n}{k}ky=ny$. Thus, $n=1$ and $y=p$ or $n=p$ and $y=1$. If $n=1$, then $p=\binom{n}{k}=1$ or $p=\binom{n}{k}=0$, independent of the value of $k$. Either way, $p$ would not be prime, and so, it must be the case that $n=p$ and $y=1$. Since $n=p$, it follows that $\binom{n-1}{k-1}=ky=k$. Hence, $\frac{(n-1)!}{(n-k)!}=k!$. We can see clearly that $k=1$ satisfies the equation. If $k\neq 1$, we have that $(n-1)(n-2)\ldots(n-k+1)=k!$. Note that the left-hand side of the equation is the product of $k-1$ consecutive integers, for which the result is $k!$. This is only possible if $(n-1)(n-2)\ldots(n-k+1)=(n-1)!$, which implies $k=n-1$.
Either way, $k=1$ or $k=n-1$. This completes the proof. $\blacksquare$
My question is on the assertion "If $k\neq 1$, we have that $(n-1)(n-2)\ldots(n-k+1)=k!$. Note that the left-hand side of the equation is the product of $k-1$ consecutive integers, for which the result is $k!$. This is only possible if $(n-1)(n-2)\ldots(n-k+1)=(n-1)!$, which implies $k=n-1$."
Is this ok to assume true? if a product of $n-1$ consecutive integers equals $n!$, then those integers should be $n,n-1,n-2,\ldots,2$ right?
 [/INST] Notice that if $k = n$, then $p = \binom{n}{k} = 1$ is not prime. So let's assume $1 < k < n-1$. Then $p > n$ that implies $\gcd(p, n) = 1$ and by the equation $kp = n\binom{n-1}{k-1}$ we have $n \mid kp$. But this implies $n \mid k$ that is impossible because $1 < k < n-1$.
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