Little math problem

For pi day here is a little mathematical proof problem for math geeks.

For every natural number n, 20 + 21 + ...+ 2n = 2n+1 − 1.

It's true for n = 1 to 5 as follows:

n=1: 20 + 21 = 3 = 22 − 1

n=2: 20 + 21 + 22 = 7 = 23 − 1

n=3: 20 + 21 + 22 + 23 = 15 = 24 − 1

n=4: 20 + 21 + 22 + 23 + 24 = 31 = 25 − 1

n=5: 20 + 21 + 22 + 23 + 24 + 25 = 63 = 26 − 1

The general formula for all n is proven by mathematical induction as follows and shown here.

Base case: Set n = 0. Then 2⁰ = 1 = 2¹ − 1.

Induction step: Show that if the equation holds for any particular n, which the above does, 
it also holds for n+1.
Let n be any natural number and 2⁰ + 2¹ + ... + 2ⁿ = 2ⁿ⁺¹ − 1 is true.

Then 2⁰ + 2¹ + ...+ 2ⁿ + 2ⁿ⁺¹ = ( 2⁰ + 2¹ + ... + 2ⁿ ) + 2ⁿ⁺¹

                                               = ( 2ⁿ⁺¹ - 1) + 2ⁿ⁺¹

                                               = 2 * 2ⁿ⁺¹ - 1

                                               = 2ⁿ⁺² - 1. QED

There is another proof as well. I will post it this weekend to give readers some time to try to find it on their own.

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March 16. The other proof follows. 2⁰ + 2¹ + ...+ 2ⁿ is a sum of a geometric progression. Using long division, the quotient of (xn+1 − 1)÷(x - 1) is xn + xn-1 + xn-2 +... 1 = xn + xn-1 + xn-2 +...+ x0.
Reverse the order of the sum and let x = 2:
20 + 21 + ... + 2n = (2n+1 − 1)÷(2 - 1) = 2n+1 − 1.