I wrote a draft of next statement for even perfect numbers that I believe that isn't in the literatute. I am asking to know a rigorous and simple proof.
Question. Prove that for each fixed even perfect number, denoted with $N=2^{p-1}(2^p-1)$, there exist two integers denoted with $A$ and $B$, in fact that those are $$A=\prod_{k=1}^p\operatorname{rad}(d_k)$$ and $$B=\prod_{k=1}^p\operatorname{rad}\left(\frac{N}{d_k}\right),$$ where $d_k$ denotes the $k$th divisor of $N$ (there are $\sigma_0(N)=2p$) $$1=d_{1}<d_2<\ldots <d_{\frac{\sigma_0(N)}{2}}=2^{p-1}<d_{1+\frac{\sigma_0(N)}{2}}=2^p-1<\ldots <d_{\sigma_0(N)}=N,$$ and $\operatorname{rad}(m)$ denotes the radical of the integer $m\geq 1$, see this Wikipedia, and such that then our even perfect number has the representation $$N=A^{1-\frac{1}{p}}B^{\frac{1}{p}}\tag{1}.$$ Many thanks.
Thus I am asking to prove rigorously the statement then.
The case $p=2$ thus $N=6$ is the first even perfect number is easy.
Thus we consider that $p>2$ and write the divisors as above, after we compute the quantity $\operatorname{rad}(d_k)$, when $1\leq k\leq \frac{\sigma_0(N)}{2}=p$ we conclude that $$A=\prod_{k=1}^p\operatorname{rad}(d_k)=1\cdot(2)^{p-1},\tag{2}$$ and similarly one has that can calculate the factors $\operatorname{rad}(N/d_k)$ as $2(2^p-1)$ if $1\leq k\leq \frac{\sigma_0(N)}{2}-1$ and $\operatorname{rad}(\frac{N}{d_{p}})=2^p-1$ thus $$B=\prod_{k=1}^p\operatorname{rad}\left(\frac{N}{d_k}\right)=(2(2^p-1))^{p-1}\cdot(2^p-1)=2^{p-1}(2^p-1)^p.\tag{3}$$ Thus from $(2)$ and $(3)$ one get $(1)$ since $$\left(\frac{B}{A}\right)^{\frac{1}{p}}=2^p-1$$ the Mersenne prime associated to the even perfect number $N$.