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.