Alternate proof for Vieta's formula (formula for the summing the roots of a polynomial) I just saw Vieta's formula for the first time, where it was stated that given some polynomial 
$$p(x)=a_nx^n+\cdots+a_0,$$
let $x_1,\ldots,x_n$ denote its roots. Then $$\sum_{i=1}^n x_i=-\frac{a_{n-1}}{a_{n}}.$$
I initially tried to use an inductive argument, starting with a linear base case, and supposing that the formula held for all polynomials of degree $m < n + 1$, and trying to move from here, by factoring out a term of $x$. This failed for what are perhaps obvious reasons.
My second attempt was successful, and I think I've found what is the standard proof, namely using the fundamental theorem of algebra to get linear factors, expanding the right side of the equation, setting the terms equal, and deriving the formula from there. This basic idea  can be found in detail at The art of problem solving.
I'm interested in an alternate proof. To clarify, I'm not looking for a proof that doesn't depend on the fundamental theorem of algebra in any way at all, but simply a different method of proving the result.
 A: If we let $r_1, .. r_n$ be the roots of $a_nx^n + .. + a_0$.
$(a_nx^n + .. + a_0)/(x - r_1) = a_nx^{n - 1} + (r_1a_n + a_{n - 1})x^{n - 2} + (r_1^2a_n + r_1a_{n - 1} + a_{n - 2})x^{n - 3} + .. + (r^{n - 2}_1a_n + .. + a_2)x + (r_1^{n - 1}a_n + .. + a_1)$
By inductive hypothesis, we have $\sum_{k = 2}^{n} r_k = -\dfrac{r_1a_n + a_{n - 1}}{a_n}$. The result falls out for $\sum_{k = 1}^{n} r_k$.
Likewise, by inductive hypothesis, $\prod_{k = 2}^n r_k = (-1)^{n - 1}\dfrac{r_1^{n - 1}a_n + .. + a_1}{a_n}.$ Then $\prod_{k = 1}^n r_k = (-1)^{n - 1}\dfrac{r_1^na_n + .. + r_1a_1}{a_n} = (-1)^{n - 1}\dfrac{-a_0}{a_n} = (-1)^n\dfrac{a_0}{a_n}$.
A: Let's appeal to combinatorics and calculus! Let's GO!
Groundwork (After this section it's smooth sailin, I promise!):
Consider a polynomial of degree $n$ factored out as so:
$$ P(x) = a (x-x_1)(x-x_2)...(x-x_n)$$
Now we introduce a notation:
$$P_q(x) =  \frac{P}{(x-x_q) }$$
Basically the notation denotes to deleting a root out of the zero set, so for example:
$$P_1 (x) = \frac{P}{x-x_1}$$
For repeated deletion, just stack on the numbers:
$$P_{123} = \frac{P}{(x-x_1)(x-x_2)(x-x_3)}$$
Note that we can shuffle the indices and it's the quantity under speaking is invariant:
$$P_{123} = P_{321} = P_{132}$$
Are all the same thing!
And one more property:
$$P_{ii} = 0$$
That is repeated number in the string makes the expression zero totally

OK! Hope that was simple enough, now it's show time, we start by observing pattern's with the derivatives of $P$:
$$ \frac{d}{dx} P = \sum_{q=1}^n P_q$$
And,
$$ \frac{d^2}{dx^2} P = \sum_{q,w=1}^n P_{qw}$$
In the above example, take for example in the summation $P_{12}$ , this is same as $P_{21}$ due to the order which we remove roots not mattering. Now we can remove the repeats by taking out the ways to permute two numbers and inducing an order on the indices (note this [1]):
$$ \frac{d^2}{dx^2} P = 2! \sum_{q \leq w , (q, w) \geq 1}^n P_{qw}$$
Now consider taking the first kth derivatives,
$$ \frac{d^k}{dx^k} P = \sum_{ \text{k indices} , ( \text{k indices} ) \geq 1}^n P_{ \text{k indices}}$$
Now by applying the arguement in [1],  we can induce an order in the summation index and factor out repeats:
$$ \frac{d^k}{dx^k} P = k! \sum_{ \text{ ordered k indices}, \text{ k indices} \geq 1 }^n P_{k indices}$$
Recall that ordered k indices greater with each index having a domain from 1 to n , is basically stating the unique ways to pick out $k$ numbers from the numbers $ \leq n$
Evaluating at $x=0$
$$ \frac{d^k}{dx^k} P = k! (-1)^{n-k} \sum_{ \text{ordered k indices } , \text{k indices} \geq 1}^n \text{ product of n-k roots} $$
By maclaurain's theorem, the polynomial expansion of $P$ has coefficent for $x^k $ as:
$$ a_k  =  \frac{1}{k!} \frac{d^k}{dx^k} P(0) = (-1)^{n-k} \sum_{ \text{ordered k indices} , \text{k indices} \geq 1}^{ n  }  \text{product of n-k roots} $$
Which is exactly the statement of Vieta's Formula
