Given a positive integer $ n> 2 $. Prove that there exists a natural number $ K $ such that for all integers $ k \ge K $ on the open interval $ \big({{k} ^{n}}, \ {{(k + 1)} ^{n}}\big) $ there are $n$ different integers, the product of which is the $n$-th power of an integer.
Source Ukrainian TST 2011
Progress: Maybe one can choose the smallest prime divisor $q$ of $n$ and then one can choose all $\frac{n}{q}$ powers and among these powers, one can choose $n$ integers whose product is a $q$-th power
In this way, we would just have to prove that
Between $k^q$ and $(k+1)^q$ we have $n$ integers with their product being a $q$-th power.
Best Answer
From the intuition that the intervals between consecutive $n$th powers contain many $(n-1)th$ powers, we have the following:
If $n+1$ is odd, for the interval in $(k^{n+1},(k+1)^{n+1})$, we can take the following $n+1$ numbers: $x_1=a^n,x_2=a^{n-1}(a+1),\ldots,x_n=a(a+1)^{n-1},x_{n+1}=(a+1)^n$, where $a=\lceil k^{\frac{n+1}{n}} \rceil$. Clearly $a^n$ falls into the interval, and what needs to be checked is that $(a+1)^n<(k+1)^{n+1}$, and we know that $a+1 < k^{\frac{n+1}{n}}+2$. We can see from binomial expansion that for fixed $n$, the first terms cancel and the leading term has degree $n$ on the RHS and degree $\dfrac{n^2-1}{n}<n$ on the LHS, so for sufficiently large $k$, all these numbers fall into the interval.