There exist $n$ different integers in the interval $\big(k^n,(k+1)^n\big)$ whose product is a perfect $n$-th power. 
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.
 A: 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.
A: Okay so finally I think this is the solution $   $
I will just be showing for even $n$
First let $a=\lceil k^{\frac{n}{n-1}} \rceil$
now let $b=a+1$ and $c=a+2$
So setting $x_{i}=a^{n-i}b^{i-1}$ and $y_{i}=c^{n-i}b^{i-1}$ for $i\in [0,n-1]$
We get  $k^n<x_{1}<\cdots x_{n-1}<y_{n-1}<\cdots y_{1}<( k^{\frac{n}{n-1}}+3)^{n-1}$ for all $k\geq 3^{n-1}$
Also note that $( k^{\frac{n}{n-1}}+3)^{n-1}<(k+1)^{n}$ for all $k\geq 3^{n-1}$
From here we just have to make cases and choose $n$ numbers out of these $2n-2$ numbers. 
For example for the case $n=4m$ the sequence $(x_{1},x_{3},\cdots, x_{4m-1},y_{1},y_{3},\cdots y_{4m-1})$ works. 
