This is one thing that I noticed. Then tried proving it, the below is the proof.

There is a proof that we can get through binary.

Take the RHS and express it in binary, The binary of

1 is 1 2 is 10 4 is 100 . .

In the same way, (2^n)-1 is 100..(n-1)0’s

So when I express the sum 1+2+4+…2^(n-1) as binary, I’ll get 1111…​ (n-1 1’s)

Now consider the RHS,

111…​(n-1 1’s) + 1

So adding 1 at the last will keeps giving a 1 carried to its preceding position. And at last you’ll get 100…​.(n 0’s)

Which when expressed in decimal is 2^n That’s it..:-)