Task is to find $f(n)$ in the following equation: $\binom {2^n} {n^4} = (f(n)+ o(1))^n$
I've found that the problem is a bit over my head. I'm attaching my partial solution below:
With use of the Stirling formula we proceed like that:
$$\binom {2^n} {n^4} = \frac { \sqrt{2\pi 2^n} \cdot {{(2^n)}^2}^n \cdot {{(1 / e)}^2}^n } { \sqrt{2\pi n^4} \cdot {{(n^4)}^n}^4 \cdot {{(1 / e)}^n}^4 \cdot \sqrt{2\pi (2^n-n^4)} \cdot (2^n-n^4)^{2^n-n^4} \cdot (1 / e)^{2^n-n^4} } = $$
$$= \frac {1 } {\sqrt{2\pi}} \frac { (2^n)^{2^n+1/2} } { {(n^4)}^{n^4 + 1/2} \cdot (2^n-n^4)^{2^n-n^4 + 1/2} }$$
Now it's kind of tempting to factor out power $n$:
$$ \sim \big( { \frac { 2^{n^2} } { (2^n-n^4)^{2-n^3} } \cdot \frac {1} {(n^4)^{n^3}} } \big)^n = \big( { \frac { 2^{n^2} } { (2^n-n^4)^{2} } } \cdot \frac {{(2^n-n^4)^{n^3}}} {(n^4)^{n^3}} \big)^n$$
Where first member approaches one, which leads me to belief that $f(n) =(\frac {2^n}{n^4})^{n^3}$.
I'm not sure that this cannot be simplified more. And also I'm not sure that it's correct. I would appreciate if you guys take a look.
This is kind of a homework, so I would be totally happy with only pointers and partial answers.