Product of $n$ consecutive positive integer is not a $n$th power? If $n>2$ and $k$ is positive integer, then there is no positive integer $m$ satisfy that $$k(k+1)\cdots (k+n-1)=m^n\, ?$$
I tried to prove this problem, but I don't know how to prove it. I know that every product of consecutive integers is not a power, but I want to simple proof of my question. Thanks for any help.
 A: I think I found a simple solution, will need someone to verify.
suppose $m$ exists such that $k(k+1)⋯(k+n−1)=m^n$
then $k < m < k + n - 1$
$gcd(m,m+1) = 1$
$m+1$ divides the LHS but doesn't divide the RHS
A: This, and variations of it appears in:
P. Erdős and J. L. Selfridge, Product of Consecutive integers is never a
power, Illinois J. of Math. 19(1975) 292- 301.
A: This solution is false. (I cannot delete an accepted answer.)

Hint: Because you have the same number of terms on both sides, $m$ must be very close to $k + \frac{n}{2} $.
The $n$ odd case is pretty easy, you should be quickly able to show that
$$(k + \frac{n-1}{2} -1 )^n < k(k+1) \ldots (k+n-1) < ( k + \frac{n-1}{2} ) ^n, $$
hence there is no integer $m$ that will work. The RHS works by pairing up terms of the form $(k+ \frac{n-1}{2} - i)(k+ \frac{n-1}{2} + i ) < (k + \frac{n-1}{2} ) ^2$, and the LHS works by pairing up terms of the form $(k+ \frac{n-1}{2} - i)(k+ \frac{n-1}{2} + i +1 ) > (k + \frac{n-1}{2} -1 )^2 $
Do the same for $n$ even, with a slightly different inequality.
