TITLE: are there known cases where $\binom{n}{k}$ is a perfect prime power?
QUESTION [15 upvotes]: I was wondering about cases where $\binom{n}{k}=p^j$ with $p$ a prime (nontrivially, so that $ n-k>1$ and $n \neq p^j$.) I had the terrible idea of checking binomial expansions
$$(x+y)^n \equiv x^n+y^n.$$
Upon some further googling, I learned that $\binom{50}{3}=140^2$, which means that this can occur for composite numbers.
Are there any ways that one can use primeness to argue that this will not occur? Or, are there any known counterexamples in which $\binom{n}{k}$ is a prime power?

Edit:
Thanks to the comments, this question answers the question for cases of $j \geq 2$ and $4 \leq k \leq n-4$. 
The main answer mentions that there are examples where $k=j=2$, are any among these solutions with $p$ a prime? 
this question gives a negative answer for $p=2$ with $n-k \geq 2$.
I suppose there is only one question left then: can the theorem be strengthened to contain the extreme cases with the assumption that $p>2$ is a prime.

REPLY [3 votes]: Erdos proved (in 1951) the stronger result that $\binom{n}{k}$ is never a perfect power if $k > 3$ (assuming that $n \geq 2k$ as we may). His argument is completely elementary -- see https://www.renyi.hu/~p_erdos/1951-05.pdf.
For $k=2$ and $k=3$, the case of possible prime powers is easily treated by divisibility arguments. The more general situation of arbitrary perfect powers for these values of $k$ was handled by Gyory in 1997 (in Acta Arithmetica), by a rather more complicated proof.