proof of Vieta's formula

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Proof: We can write $P(x)$ as

P(x)=a_{n}(x-r_{1})(x-r_{2})\ldots(x-r_{n})=a_{n}(x^{n}-S_{1}x^{{n-1}}+S_{2}x^% {{n-2}}-\ldots+(-1)^{n}S_{n}).

Since we also have that

P(x)=a_{n}x^{n}+a_{{n-1}}x^{{n-1}}+\ldots a_{{1}}x+a_{{0}},

setting the coefficients equal yields

a_{n}(-1)^{{i}}S_{i}=a_{{n-i}}

which is what the theorem stated.

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