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Suppose I want to show that if an $n$ element set has $2^n$ subsets, then $\sum_{k=0}^n \binom{n}{k}=2^n$

How would I go by doing this without using any theorems?

Anyway thanks.

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    Can you use the fact that your set has exactly $\dbinom{n}{k}$ subsets of size $k$? – 2017-01-14
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    Usually the statement is proven the other way around, but if you count all of the subsets directly as "# of 0-element subsets + # of 1-element subsets + # of 2-element subsets + ..." then this becomes obvious (*depending on your definition or understanding of $\binom{n}{k}$*) – 2017-01-14
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    @AndreasCaranti Yes. How would I go by doing it? – 2017-01-15
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    As an aside, the usual exercise is to start with $\sum\limits_{k=0}^n\binom{n}{k}$ and prove that it equals $2^n$ first without the use of the hypothesis in your problem, and use this to conclude that an $n$ element set has $2^n$ subsets. (*approach via binomial theorem for $(1+1)^n$*) – 2017-01-15
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    Could I show that there are an $\binom{n}{k}$ subsets of size $k$ given that there are $2^n$ subsets in an $n$-element set? – 2017-01-15

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We need to find how many subsets of size $0,1,2,\dots,n$ has a set of $n$ elements. Take care that the order of the selection doesn't matter.

$\mathbf{0}$ elements: In how many ways can I take $0$ elements from a set of $n$ elements? $\binom{n}{0}$
$\mathbf{1}$ element: In how many ways can I take $1$ element from a set of $n$ elements? $\binom{n}{1}$
$\mathbf{2}$ elements: In how many ways can I take $2$ elements from a set of $n$ elements? $\binom{n}{0}$
$\vdots$
$\mathbf{n}$ elements: In how many ways can I take $n$ elements from a set of $n$ elements? $\binom{n}{n}$

Then a set of $n$ elements has $\displaystyle\sum_{i=0}^{n}\binom{n}{i}$ subsets.

Note that $\displaystyle(1+1)^n=\sum_{i=0}^{n}\binom{n}{i}\Rightarrow2^n=\sum_{i=0}^{n}\binom{n}{i}$.