Finding Big-O of the following function I have the following question that I need to give the Big Oh estimate for:
$$(n^n +n2^n+5^n)(n!+2^n).$$
I tried to expand this:
$$n^nn! + n^n2^n + n2^nn! + n2^{2n} + 5^nn! + 5^n2^n$$
Would the Big Oh then be $O(n^nn!)$?

For any function $f(n)$ where $$(n^n +n2^n+5^n)(n!+2^n) < Cf(n)$$ for all $n$, where $C>0$ is a constant, you can say your function is $O(f(n))$.
In this case, yes your function is $O(n^n n!)$. As another commenter pointed out, you can estimate $n!$ by $\left(\frac{n}{e}\right)^n \sqrt{n}$, up to a constant, so your function is $O\left(\frac{n^{2n+{1\over 2}}}{e^n}\right)$.
Maybe the neatest way to write this is
$$O\left(\left(\frac{n^2}{e}\right)^n \sqrt{n}\right).$$
