Why $2^{n\log_2n}=n^n$? I couldn't find any answer on the internet so I came here, I'm trying to understand how $2^{n\log_2n}=n^n$
I know that $n^n=e^{\ln{n^n}}=e^{n\ln{n}}$
So far I've got $2^{n\log_2n}=2^{\log_2{n^n}}=2^{\frac{\ln{n^n}}{\ln2}}$
Then I'm stuck, I'm not even sure I'm going the good way..
Thanks in advance
 A: $\log_2(x)$ is defined to be the number $y$ such that $2^y = x$.
Therefore $$2^{n \log_2(n)} = \left(2^{\log_2(n)}\right)^n = n^n$$
Your way will work, but you need to write the bottom $2$ as $e^{\ln(2)}$ before you can make progress.
A: Note that by definition
$$x=\log_2^n\iff 2^x=n$$
thus
$$2^{n\log_2n}=(2^x)^n=n^n$$
A: $$2^{n\log2^n}=2^{\log2^{n^n}}$$
$\log 2$ and $2$ are inverses, so they cancel out, so 
$$2^{n\log2^n}=n^n$$
Let me explain why they cancel out.
You know $\log_a a^x=x$, and you need to prove $a^{\log_a x}=x$
Set $\log_ax=y$, and make this equation in exponential form, so $\log_ax=y \implies x=a^y$.
Using substitution, $a^{\log_ax}=a^{\log_aa^y}$. Using what you know, $a^{\log_aa^y}=a^y$.
Since $a^y=x$, $a^{\log_a x}=x$.
Use this to help you prove $2^{n \log_2n}=n^n$
A: For the same reason as $\; n^n=\mathrm e^{n\log n}$. Only the basis of the logarithm changes.
A: taking the logarithm on both sides we get
$$n\log_{2}{n}\ln(2)=n \ln(n)$$ dividing by $$\ln(2)$$ we obtain
4$$n\log_{2}{n}=n\frac{\ln(n)}{\ln(2)}=n\log_{2}{n}$$ which is true.
