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While I was calculating summations of combinations I realized that $2^x = \sum^x_{n=0}\binom{x}{n}$ for all positive integers $x$. Surprisingly, there are no exponents in the definition of combination and I am curious that why this relation holds.

From a geometric point of view, I can see how they are both describing the number of corners in a "cube" in $x$-dimension. Can it be proven in a more mathematical approach?

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    How many subsets of $\{1,2,3,\dots,x\}$ exist? How many subsets of size $n$? – 2017-02-10
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    This is also immediate from the binomial theorem $(a+b)^x = \sum\limits_{n=0}^x \binom{x}{n}a^nb^{x-n}$ (*trying to mimic your notation*) by setting $a=b=1$. – 2017-02-10
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    @JMoravitz You must have been tempted to answer, "Because 1+1=2." – 2017-02-10
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    Compare this question with http://math.stackexchange.com/questions/652619/pascals-triangle-proof or http://math.stackexchange.com/questions/1375856/why-does-n-choose-r-where-r-1-n-track-2n – 2017-02-10
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    @DavidK Coming from a CS background, [this answer](http://math.stackexchange.com/a/1375906/33706) in your link makes the most sense to me. Thanks for the link! – 2017-02-10

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There is a tricky and simple proof for it: $$2^x=(1+1)^x=\sum_{n=0}^x \binom{x}{n}1^x1^{n-x}=\sum_{n=0}^x \binom{x}{n}$$