Use the change-of-base identity and the exponent property for logarithms.
[tex]\log_b(a) = \dfrac{\log_c(a)}{\log_c(b)} \implies \log_b(a) = \dfrac1{\log_a(b)}[/tex]
[tex]\log_c(a^b) = b \log_c(a)[/tex]
This lets us rewrite
[tex]\log_{(k+1)^{2k-1}} (k+2)^{2k} = \dfrac{2k}{\log_{k+2}(k+1)^{2k-1}} = \dfrac{2k}{(2k-1) \log_{k+2}(k+1)}[/tex]
[tex]\log_{(k+2)^{2k-1}} (k+1)^{k^2} = \dfrac{k^2}{\log_{k+1} (k+2)^{2k-1}} = \dfrac{k^2}{(2k-1) \log_{k+1}(k+2)}[/tex]
It follows that
[tex]a_k = \dfrac2k \dfrac{\log_{k+2}(k+1)}{\log_{k+1}(k+2)} = \dfrac2k \left(\ln^2(k+1)}{\ln^2(k+2)}[/tex]
so that the product of [tex]a_k[/tex] telescopes:
[tex]\displaystyle \prod_{k=1}^{n-1} a_k = \frac{2^{n-1}}{(n-1)!} \frac{\ln^2(2)}{\ln^2(3)}\cdot\frac{\ln^2(3)}{\ln^2(4)}\cdot\frac{\ln^2(4)}{\ln^2(5)} \cdots \frac{\ln^2(n-1)}{\ln^2(n)} \cdot \frac{\ln^2(n)}{\ln^2(n+1)} \\\\ ~~~~~~~~ = \frac{2^{n-1}}{(n-1)!} \frac{\ln^2(n+1)}{\ln^2(2)}[/tex]
We can similar reduce the coefficient of [tex]P_n[/tex] to write
[tex]\log_{(n+1)^{2n-1}}(2^{2n}) = \dfrac{2n}{(2n-1) \log_2(n+1)}[/tex]
[tex]\log_{2^{2n-1}}(n+1)^{n^2} = \dfrac{n^2}{(2n-1)\log_{n+1}(2)}[/tex]
[tex]\implies \dfrac{\log_{(n+1)^{2n-1}}(2^{2n})}{\log_{2^{2n-1}}(n+1)^{n^2}} = \dfrac2n \dfrac{\ln^2(2)}{\ln^2(n+1)}[/tex]
Then the expression for [tex]P_n[/tex] reduces significantly to
[tex]P_n = \dfrac2n \dfrac{\ln^2(2)}{\ln^2(n+1)} \cdot \dfrac{2^{n-1}}{(n-1)!} \dfrac{\ln^2(n+1)}{\ln^2(2)} = \dfrac{2^n}{n!}[/tex]
and we ultimately find
[tex]\displaystyle \sum_{n=2}^\infty P_n = \sum_{n=0}^\infty \frac{2^n}{n!} - \frac{2^1}{1!} - \frac{2^0}{0!} = \boxed{e^2 - 3}[/tex]
where we recall the power series
[tex]\displaystyle e^x = \sum_{n=0}^\infty \frac{x^n}{n!}[/tex]
