$$f(n)=\Bigg(1-\Big(1-\frac{1}{2^{n/2}}\Big)^n\Bigg)^{n^7}$$
I am interested in large $n$. The number $7$ can be replaced by any fixed integer.
I have $$f(n)\rightarrow\Bigg(1-e^{-n2^{-n/2}}\Bigg)^{n^7}\rightarrow e^{-e^{-n2^{-n/2}}n^7}$$
$$f(n)=\Bigg(1-\Big(1-\frac{1}{2^{n/2}}\Big)^n\Bigg)^{n^7}$$
I am interested in large $n$. The number $7$ can be replaced by any fixed integer.
I have $$f(n)\rightarrow\Bigg(1-e^{-n2^{-n/2}}\Bigg)^{n^7}\rightarrow e^{-e^{-n2^{-n/2}}n^7}$$
For large $n$, $n2^{-n/2}<<1$ so the $e^{-x}\approx 1-x$ approximation works. Let's replace $7$ with $k$. We have $$\ln f = n^k\ln\left(1-\left(1-2^{-n/2}\right)^n\right)\approx n^k\ln\left(n2^{-n/2}\right)=n^k\left(-n\ln\sqrt{2}+\ln n\right).$$Hence$$f\approx n^{n^k}\sqrt{2}^{-n^{k+1}}.$$