<s>[INST] Why is it that $2^{10} + 2^{9} + 2^{8} + \cdots + 2^{3} + 2^2 + 2^1 = 2^{11} - 2$? Why is it that $$2^{10} + 2^{9} + 2^{8} + \cdots + 2^{3} + 2^2 + 2^1 = 2^{11} - 2?$$
 [/INST] Here's an alternative argument using the fact that a set of $n$ elements has $2^n$ subsets.  Consider subsets $S$ of the set $\{0,1,2,\dots,9,10\}$.  Since this set has 11 elements, it has $2^{11}$ subsets. Now ask, for each $n$ in the range $0\leq n\leq 10$, how many of these subsets $S$ have $n$ as their largest element.  Well, any such $S$ consists of the element $n$ together with some subset $S'$ of $\{0,1,\dots,n-1\}$.  So there are $2^n$ choices for $S'$.  Thus, $n$ is the largest element of exactly $2^n$ subsets $S$ of $\{0,1,2,\dots,9,10\}$.  Therefore, the sum $2^{10}+2^9+\dots+2^2+2^1$ counts all the subsets $S$ of $\{0,1,2,\dots,9,10\}$ whose largest element is 10 or 9 or $\dots$ or 2 or 1.  That means it counts all of the $2^{11}$ subsets $S$ of $\{0,1,2,\dots,9,10\}$ except for two, namely the empty set and the set $\{0\}$.  So the sum is $2^{11}-2$.
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