# Why is $log(n)+log(\frac{n}{2})+log(\frac{n}{4})+log(\frac{n}{8})+log(\frac{n}{16})+…+log(\frac{n}{n})=\Theta (log^2(n)$

$$log(n)+log(\frac{n}{2})+log(\frac{n}{4})+log(\frac{n}{8})+log(\frac{n}{16})+…+log(\frac{n}{n})=\Theta (log^2(n)$$

The sum of logarithms is the logarithm of the product $$n*\frac{n}{2}*\frac{n}{4}*\frac{n}{8}*\frac{n}{16}*…*\frac{n}{n}$$. This equals $$n^{log(n)}$$ divided by what? If the product would just be $$n^{log(n)}$$, then this would make perfect sende since $$log(n^{log(n)})=log(n)*log(n)=log^2(n)$$. But the divisor equals $$\frac{1}{2}*\frac{1}{4}*\frac{1}{8}*\frac{1}{16}*…*\frac{1}{n}$$ so I don’t get it.