# 1/(2^n)! is negligible function?

By definition $$\frac {1} {n}$$, $$\frac{1}{2^n}$$ and $$\frac{1}{n!}$$ are negligible functions.

I have got the function $$f(n) = \frac{1}{(2^n)!}$$ where $$n$$ is security parameter.

I don't understand, How do I formally proof that $$\frac{1}{(2^n)!}$$ is a negligible function? Can anyone please help me out?

The simplest way to prove a function $$f$$ is negligible if it is obviously negligible is to show that it is "more negligible" than some other function $$g$$ which you have already proven negligble, e.g. $$g(n)=2^{-n}$$.
Because $$g$$ is negligble, there exists $$n_{g_0}$$ such that for all $$n>n_{g_0}$$ it holds that $$g(n)<1/{n^c}$$ for any fixed choice of $$c$$.
Now you can capitalize on that by showing that there exists some $$n_{f_0}$$ such that for all $$n>n_{f_0}$$ it holds that $$f(n)\leq g(n)$$. Clearly it then holds that for all $$n>\max(n_{f_0},n_{g_0})$$ that $$f(n)\leq g(n)<1/n^c$$ which means $$f$$ is negligible.
I'll leave the choice of $$g$$ for your concrete application to you.