Binomial coefficients that are powers of 2 I would like a proof that
$$ {{n}\choose{k}} = \frac{n!}{k!(n-k)!} = 2^m $$
for $n,k,m\in \mathbb{N}$, only if $k=1$ or $k=n-1$.
It seems to me that this must be true since for other values of $k$ the numerator contains more factors that are not powers of 2 than the denominator. Furthermore, the numerator also contains larger factors than the denominator and thus can't all be cancelled. Nevertheless, I have been unable to form an elegant, watertight proof.

Betrand's Postulate implies that for $n \ge 1$ there is always a prime $p$ with $n < p \le 2n$.
Sylvester strengthened this result to:

If $n \ge 2k$ then at least one of the numbers $n, n - 1, n - 2, \cdots, n - k + 1$ has a prime divisor $p > k$.

Hence if $n \ge 2k$, which we can always assume since $\displaystyle \binom{n}{k} = \binom{n}{n - k}$, then
$$ \binom{n}{k} = \frac{n(n-1)\cdots(n-k+1)}{k!} $$
has a prime $p$ in the numerator with $p > k$. Hence $p \mid \binom{n}{k}$. Since we are assuming $k, n - k \ge 2$ this shows that $\binom{n}{k}$ is not a power of $2$.
Reference
M. Aigner Proofs from THE BOOK, Springer (2014), 5th ed. (Ch. 2, 3)
