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..:-)

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