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I just relized, I have no idea how to evaluate:

$$n2^2n!^2$$

Google didn't help me find a breakdown could anybody just space it out?

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$$ 2^2n(n!)^2=4n(n!)(n!)=4n\left(n\cdots 2\cdot 1\right)\left(n\cdots 2\cdot 1\right)=4n\left(n^2\cdots 2^2\cdot 1^2\right) =4n\prod_{k=1}^n k^2 $$

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If you want an approximate value for a large value of $n$, you can use Stirling approximation $$n! \approx \sqrt{2 \pi n}\Big(\frac{n}{e}\Big)^n$$ and so $$n2^2n!^2=4n(n!)^2\approx 8\pi n^2\Big(\frac{n}{e}\Big)^{2n}$$ For example, if $n=100$, the exact value is $3.484\times 10^{318}$ while the approximation is $3.478\times 10^{318}$ ; if $n=1000$, the exact value is $6.477\times 10^{5138}$ while the approximation is $6.476\times 10^{5138}$.

For sure, there better (but more complex) approximations for $n!$ that you could use.

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Think of it like this:

$$n \cdot (2^2)\cdot [(n!)^2] \\ =n \cdot 4 \cdot(n!)(n!)$$

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