Theorem: Product of Roots of Polynomial

Let $P$ be the polynomial equation:
:$a_n z^n + a_{n - 1} z^{n - 1} + \cdots + a_1 z + a_0 = 0$
such that $a_n \ne 0$.
The product of the roots of $P$ is $\dfrac {\paren {-1}^n a_0} {a_n}$.

Proof:
Proof by induction.
For $n \in \N_{> 0}$, let $\map S n$ be the statement:
:For polynomial $P$ of degree $n$, the product of the roots of $P$ is $\dfrac {\paren {-1}^n a_0} {a_n}$.
=== Basis of Induction ===
Let $P$ be the polynomial of degree $1$:
:$\map P x = a_0 + a_1 z$
The single root of $P$ is:
:$a_0 + a_1 z = 0 \implies z = -\dfrac {a_0} {a_1}$
So, $\map S 1$ is true.
{{qed|lemma}}
=== Induction Hypothesis ===
Now we must show that $\map S k \implies \map S {k + 1}$ for all $k \geq 1$.
That is, we assume the induction hypothesis:
:For polynomial $P$ of degree $k$, the product of the roots of $P$ is $-\dfrac {\paren {-1}^k a_0} {a_k}$.
in order to prove that:
:For polynomial $P$ of degree $k + 1$, the product of the roots of $P$ is $-\dfrac {\paren {-1}^{k + 1} a_0} {a_{k + 1} }$.
=== Induction Step ===
Let $P$ be a polynomial of degree $k + 1$.
Let $\lambda$ be a root of $P$.
Then, by the Polynomial Factor Theorem, $P$ can be written as:
:$\map P z = \paren {z - \lambda} \map Q x$
where $Q$ is a polynomial of degree $k$.
Let $Q$ be written as:
:$\ds \sum_{i \mathop = 0}^k b_i z^i$
We can now write $\map P z$ as
{{begin-eqn}}
{{eqn | l = \map P z
      | r = \paren {z - \lambda} \map Q x 
}}
{{eqn | r = \paren {z - \lambda} \sum_{i \mathop = 0}^k b_i z^i
}}
{{eqn | r = z \sum_{i \mathop = 0}^k b_i z^i - \lambda \sum_{i \mathop = 0}^k b_i z^i
      | c = Multiplication Distributes over Addition
}}
{{eqn | r = \sum_{i \mathop = 0}^k b_i z^{i + 1} - \sum_{i \mathop = 0}^k \lambda b_i z^i
      | c = Indexed Summation of Multiple of Mapping
}}
{{eqn | r = b_k z^{k + 1} + \sum_{i \mathop = 0}^{k - 1} b_i z^{i + 1} - \sum_{i \mathop = 1}^k \lambda b_i z^i - \lambda b_0
}}
{{eqn | r = b_k z^{k + 1} + \sum_{i \mathop = 1}^k b_{i - 1} z^i - \sum_{i \mathop = 1}^k \lambda b_i z^i - \lambda b_0
      | c = {{Corollary|Translation of Index Variable of Summation}}
}}
{{eqn | r = b_k z^{k + 1} + \sum_{i \mathop = 1}^k \paren {b_{i - 1} z^i - \lambda b_i z^i} - \lambda b_0
      | c = Indexed Summation of Sum of Mappings
}}
{{eqn | r = b_k z^{k + 1} + \sum_{i \mathop = 1}^k \paren {b_{i - 1} - \lambda b_i} z^i - \lambda b_0
      | c = Multiplication Distributes over Addition
}}
{{end-eqn}}
So, by matching up exponents of $z$:
:$a_{k + 1} = b_k$
and:
:$a_0 = -\lambda b_0$.
Therefore:
{{begin-eqn}}
{{eqn | l = \frac {\paren {-1}^{k + 1} a_0} {a_{k + 1} }
      | r = \frac {\paren {-1}^{k + 1} \paren {- \lambda b_0} } {b_k}
}}
{{eqn | r = \lambda \frac {\paren {-1}^{k + 2} b_0} {b_k}
}}
{{eqn | r = \lambda \frac {\paren {-1}^k b_0} {b_k}
      | c = $\paren {-1}^2 = 1$
}}
{{end-eqn}}
By the induction hypothesis, the product of the roots of $Q$ is $\dfrac {\paren {-1}^k b_0} {b_k}$.
Since every root of $Q$ is a root of $P$, the final expression is the product of the roots of $P$.
So, the product of the roots of $P$ is $\dfrac {\paren {-1}^{k + 1} a_0} {a_{k + 1} }$
{{qed|lemma}}
Thus, $\map S n$ holds for all $n \in \N_{> 0}$.
{{qed}}