Find closed form of Wallis's product type $\prod_{n=1}^{\infty}\left(\prod_{k=0}^{m}(n+k)^{{m\choose k}(-1)^k}\right)^{(-1)^n}$ Wallis's product

$${2\over 1}\cdot{2\over 3}\cdot{4\over 3}\cdot{4\over 5}\cdots={\pi\over 2}\tag1$$

Generalised of Wallis's product type

$$\prod_{n=1}^{\infty}\left(\prod_{k=0}^{m}(n+k)^{{m\choose k}(-1)^k}\right)^{(-1)^n}\tag2$$
  Where $m\ge 1$

Setting $m=1$, we get $(1)$
$$\prod_{n=1}^{\infty}\left({n\over n+1}\right)^{(-1)^n}={\pi\over 
2}\tag3$$
and $m=2$ we got 
$$\prod_{n=1}^{\infty}\left(n\cdot{n+2\over (n+1)^2}\right)^{(-1)^n}={\pi^2\over 8}\tag4$$
How can we find the closed form for $(2)$? Conjecture closed form may take the form of $${\pi^{2^{m-1}}\over F(m)}$$

$$ \log P(m) = \sum_{n\geq 1}(-1)^n\sum_{k=0}^{m}(-1)^k \binom{m}{k}\log(n+k) $$
can be written, by exploiting the integral representation for the logarithm given by Frullani's theorem, as
$$ \log P(m) = \sum_{n\geq 1}(-1)^{n+1} \int_{0}^{+\infty}\frac{e^{-nx}(1-e^{-x})^m}{x}\,dx =\int_{0}^{+\infty}\frac{\left(1-e^{-x}\right)^m}{\left(1+e^x\right) x}\,dx$$
and always by Frullani's theorem the RHS of the last line is a linear combination of $\log\frac{\pi}{2}$ and logarithms of natural numbers: it is enough to reduce $(1-t)^m$ $\pmod{t+1}$, since:
$$ \log P(m) = -\int_{0}^{1}\frac{(1-t)^m}{(1+t)\log(t)}\,dt.$$
