What's an intuitive way to compute summation of this series?

What's an intuitive way to compute

$\mathrm{log}(1)+\mathrm{log}(2)+\mathrm{log}(3)+\cdots +\mathrm{log}(n-1)+\mathrm{log}(n)$

or

for $n>a$

$\mathrm{log}(a)+\mathrm{log}(a+1)+\mathrm{log}(a+2)+\cdots +\mathrm{log}(n-1)+\mathrm{log}(n)$

I know the answer for general case is

$\mathrm{log}\left(\frac{n!}{(a-1)!}\right)$

What's an intuitive way to compute

$\mathrm{log}(1)+\mathrm{log}(2)+\mathrm{log}(3)+\cdots +\mathrm{log}(n-1)+\mathrm{log}(n)$

or

for $n>a$

$\mathrm{log}(a)+\mathrm{log}(a+1)+\mathrm{log}(a+2)+\cdots +\mathrm{log}(n-1)+\mathrm{log}(n)$

I know the answer for general case is

$\mathrm{log}\left(\frac{n!}{(a-1)!}\right)$