Empty product and Vieta's formulas Following on from another simple question about Vieta's formulas which I asked here, I just want to formally confirm a conclusion which seems to be borne out by the evidence.
One of Vieta's formulas links the roots $r_i$ of a polynomial $a_n x^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ to its coefficients as follows:
NOTE THAT IN MY ORIGINAL POST, I ERRONEOUSLY PUT $\prod_{i=0}^n$
$$\prod_{i=1}^n r_i=(-1)^n\frac{a_0}{a_n}$$
I've run a few examples, and it seems that any instances of $r_i=0$ must be taken to be the empty product $1$ rather than literally $0$. Can someone confirm that this is formally the case?
 A: The definition of the empty product does not apply here. From the non-quoted part at the end of your comment ($\prod 0 = 1$) it seems you are misunderstanding what it means. 
Note that nowhere does the definition talk about any factors being $0$. It talks about the case where any product defined with the product symbol has no factors.
For example, a definition of the factorial of a positive integer $n$ can be given as
$$n! = \prod_{i=1}^n i.$$
If you choose to see what happens if you plug in $n=0$ at the right hand side (which is not in the original domain), you get $\prod_{i=1}^0 i$, which is the empty product (it has no factors, the first factor would be $1$, but it also ends at $0$, so again, no factor).
Using the definition of the empty product, we see that $\prod_{i=1}^0 i = 1$, and since that is what we want $0!$ to be, we can use the above definition also for $n=0$, so it now applies to all non-negative integers. 
The empty product (like the empty sum) is a convention to allow certain formulas to be true in edge cases, allowing definitions and statements to be made without always mentioning the edge cases as special cases that need special handling. 
Back to your formula. First, you made a small error, the product index should start with $i=1$, not $i=0$ (this happened already in your previous question), as a non-constant polynomial has at most as many roots as its degree . So the formula should be
$$\prod_{\color{red}{i=1}}^n r_i = (-1)^n\frac{a_0}{a_n}.$$
For $n=0$, there is an empty product on the left hand side. However, it would mean that you are looking at a constant polynomial that is just $a_0$. That either has no roots (if $a_0 \neq 0$), in which case Vieta's formula still works: The left hand side is an empty product, so equal to $1$. The right hand side is also $1$, as $n=0$ and $a_n=a_0$.
But if $n=0$ and $a_0=0$, then the polynomial has every real number as root, so Vieta's formula breaks down totally.
If $n>0$, then the left hand side is not an empty product, it will contain exactly $n$ factors $r_i$. If one of them is $0$, the product is $0$. And that's consistent: If $0$ is a root, putting $x=0$ into the polynomial should produce $0$. But at $x=0$, the polynomial is just $a_0$, so we get $a_0=0$ in this case, and so Vieta's formula gives a zero on both sides.
A: Let $P(x)=a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$ with $n\neq0$.
If for some index $1\leq i \leq n$, $r_i = 0$ is a root of the polynomial then $P(r_i)=0 \Rightarrow P(0)=a_0=0$.
Thus, $\displaystyle \prod_{k=1}^n r_k = r_1 \dots r_i \dots r_n = r_1 \dots 0 \dots r_n = 0 = (-1)^n \frac{a_0}{a_n}$ and there is nothing wrong. 
Also note that there is no "empty product" as we multiply all the roots since we have roots!
However, if $n=0$ then $P(x)=a_0$ and for $a_0=0$, $P(x) \equiv 0$. That is, $P$ is the zero polynomial. So, $P$ has any "infinitely many" roots and the formula clearly can not be applied. Please note that we can not even write $\prod_{k=1}^\infty r_k$ since there are "uncountable" many roots in fact! 
In additon, $(-1)^n\frac{a_0}{a_n} = (-1)^0\frac{a_0}{a_0} = \frac{0}{0}$ is also meaningless (undefined).
Besides, if $a_0 \neq 0$, there is no root. This is the case where "empty product" takes place. Now the formula looks like (note that $a_n = a_0$ also)
$\prod_{k=1}^0 r_k = (-1)^0\frac{a_0}{a_0}$
The right hand side is okay and equals to 1 but the left hand side is meaningless as the upper limit is less than the lower limit. This is because we don't we have any roots, there is nothing to multiply! Such cases are called "empty product" and for convenience they are "defined" as multiplicative identity 1. That is, $\prod_{k=1}^0 r_k := 1$. Now the formula holds even in this case (no roots case).
Similar situations occur for summation, too. When there is nothing to sum, that is called "empty sum". For example, $\sum_{k=1}^0 r_k$ is an empty sum and "defined as" additive identity 0. That is $\sum_{k=1}^0 r_k := 0$.
Note: Please feel free to edit this answer.
