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According to the binomoial theorm $$(x+y)^{n} = \sum_{j=0}^{n} \binom{n}{j}x^{n-j}y^{j}=\binom{n}{0}x^{n}+\binom{n}{1}x^{n-1}y + ... + \binom{n}{n-1}x y^{n-1}+\binom{n}{n}y^n$$

but my textbook says $$0=0^{n}=((-1)+1)^{n} = \sum_{k=0}^{n}\binom{n}{k}(-1)^{n-k}1^{k}$$

shouldn't $$(-1)^{k} 1^{n-k}$$ be $$(-1)^{n-k}1^{k}$$?

Since different k would determine different signs, namely + or -. Please fill me in. thanks.

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    I think there's a typo in the formula, one $n$ should be over the sigma, but appears to multiply $\binom{n}{k}$. I think there's a second typo where you've already switched $(-1)^k1^{n-k}$ with $(-1)^{n-k}1^k$. To answer what I think the actual question is, note that $\binom{n}{k}=\binom{n}{n-k}$ – 2017-01-20

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Hint: The representation is valid, since \begin{align*} (x+y)^n=(y+x)^n \end{align*}