proof of Vieta’s formula

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.