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I know it has something to do with the binomial theorem but I'm not sure how the changed summation limits affect it.

$$\sum_{k=1}^{n}\binom{n}{k}=\sum_{k=0}^{n-1}2^{k}$$

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$$ \sum_{k=1}^n{n\choose k}=-1+\sum_{k=0}^n{n\choose k}=-1+2^n=\frac{1-2^n}{1-2}=\sum_{k=0}^{n-1}2^k. $$

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  • $\begingroup$ Just a note to the OP. Mercy uses properties of binomial sums and geometric series to get the result. $\endgroup$ – Ali Caglayan Feb 2 '14 at 0:05
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$$(1+x)^n = \sum_{k=0}^n \binom{n}{k} x^k = 1 + \sum_{k=1}^n \binom{n}{k}x^k \implies (1+x)^n\bigg|_{x=1} = \cdots$$

$$\sum_{k=0}^{n-1} 2^k = \frac{2^n - 1}{2-1} $$

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Binomial Theorem $$ (a+b)^n=\sum_{k=0}^n \binom{n}{k}a^kb^{n-k}. $$ Set $a=b=1$ and get $$ 2^n=(1+1)^n=\sum_{k=0}^n\binom{n}{k}. $$

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$$\sum_{k=1}^{n}\binom{n}{k}=-1+\binom{n}{0}+\sum_{k=1}^{n-1}2^{k}=-1+\sum_{k=0}^{n-1}2^{k}=-1+2^n=$$ $$=2^n-1=\frac{2^n-1}{2-1}=1+2+...+2^{n-1}=\sum_{k=0}^{n-1}2^{k}$$

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