Bonjour,
The equation $\binom{n}{k}=m^l$ has no entire solution for l$\ge$2 and 4$\le$k$\le$n-4. Suppose that n$\ge$2k (since $\binom{n}{k}=\binom{n}{n-k}$).
According to the Sylvester theorem, the binomial coefficient (for n$\ge$2k):
\begin{equation} \binom{n}{k}=\frac{n(n-1)...(n-k+1)}{k!}, \end{equation}
has always a prime factor p greater than k.
I dont't know why this fact implies that $p^l$ divides $n(n-1)...(n-k+1)$.
And I dont't understand why only one of the factors n-i can be a multiple of p.
Can someone please help ?