Vieta's formula

Prerequisite knowledge:
Elementary symmetric functions:
$\sigma_k$ of n variables is the sum of all products of k variable(s)

For two variables x, y:
$\sigma_1 = x + y\\
\sigma_2 = xy$

For three variables x, y, z:
$\sigma_1 = x + y + z\\
\sigma_2 = xy + xz + yz\\
\sigma_3 = xyz$

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From our knowledge of secondary school maths on quadratic functions, we know that for a quadratic equation $ax^2+bx+c$, the sum of roots is given by -b/a whereas the product of roots is c/a.
This is easily proved as follows:
Let p, q be the two roots.
$ax^2+bx+c=0\\
x^2+\frac{bx}{a}+\frac{c}{a}=0 \:[1]\\
(x-p)(x-q)=0\\
x^2-(p+q)x+pq=0 \:[2]$
Comparing [1] and [2], we get the desired result.

In fact, this is related to Vieta's formula, which states that
$\sigma_k = (-1)^k \cdot \frac{a_{n-k}}{a_n}$ for $1\leq k\leq n$.
Proof by MI:
When $n=1$ (degree 1), $f(x)=a_{1}x+a_0$, with one root $-\frac{a_0}{a_1}$.
$\sigma_1=(-1)^1 \cdot \frac{a_0}{a_1}=-\frac{a_0}{a_1}$, as desired.
Assume the formula is true for $n=k$.
When $n=k+1$, $f(x)=a_{k+1}x^{k+1}+a_{k}x^{k}+...+a_0$
Let $g(x)$ be a monic polynomial (coefficient of degree being 1) with degree $k$ (To use our assumption) such that
$f(x)=a_{k+1}(x-r_{k+1})g(x)$, where $r_{k+1}$ is one of the $k+1$ roots of $f(x)=0$.
Then $g(x)=x^k-\sigma_1 x^{k-1}+\sigma_2 x^{k-2}-...+(-1)^k \sigma_k$
Thus $f(x)=a_{k+1}(x-r_{k+1})(x^k-(r_1+r_2+...+r_k)x^{k-1}+...+(-1)^k (r_{1}r_{2}\ldots r_{k}))$
$f(x)=a_{k+1}(x^{k+1}-(r_1+r_2+...+r_k+r_{k+1})x^{k-1}+...+(-1)^k (r_{1}r_{2}\ldots r_{k}r_{k+1}))$
$f(x)=a_{k+1}(x^{k+1}-\sigma_1 x^{k}+...+(-1)^{k+1} \sigma_{k+1}) \: \Box$

Examples:
[later]