There are complex numbers $z_{1}, \cdots z_{n}$ such that $z^{n} + \cdots a_{0} = \prod_{i=1}^{n}(z-z_{i})$ - Final induction step The following is a question from Spivak's Calculus - Ch 26 (3a)
If $a_{0}, \cdots a_{n-1}$ are any complex numbers, then there are complex numbers $z_{1}, \cdots z_{n}$ such that $$z^{n} + \cdots a_{0} = \prod_{i=1}^{n}(z-z_{i})$$
I had a question on obtaining the final conclusion from my inductive step.
This statement can be proved by induction using the fundamental theorem of algebra along with a fact that was established earlier in the book that stated

For any polynomial function $f$ and any $a$, there exists a polynomial $g$ and a constant $b$ s.t. $f(x) = (x-a)g(x) + b$

Base Case:
By the FTA we know there exists a root $z_{1}$ such that
$$(z-z_{1})(z^{n-1} + b_{n-2}z^{n-2}+ \cdots b_{1}z + b_{0})$$
Induction Hypothesis:
Suppose this is true for $n=k$, thus there exists a polynomial g(z) and constant $\gamma$ such that
$$\prod_{i=1}^{k}(z-z_{i})g(z) + \gamma$$
Induction Step:
Here is where I'm stuck. I know by the FTA there exists a root call it $z_{k+1}$ and constant $\phi$ which would allow me to write my expression as
$$\prod_{i=1}^{k+1}(z-z_{i})g(z) + \phi$$
But how would I state that this would be the end of the division algorithm that got me to this point and as such should be
$$\prod_{i=1}^{k+1}(z-z_{i})$$
?
 A: Here is how you can set out the induction if you can use the fundamental theorem of algebra and the division algorithm.
Base case
Let $\mathscr S_k$ be the statement that for any polynomial $f(z) = z^k+ a_{k-1}z^{k-1}+\cdots a_0 $ of degree $k > 0$ there exists roots $z_1, z_2, \cdots z_k$ such that
$$
f(z) = (z-z_1)(z-z_2)\cdots(z-z_k)
$$
Plainly $\mathscr S_1$ is true.
Inductive step
Suppose you have proved $\mathscr S_k$ for some $ k >0$.  Let $p(z) = z^{k+1} + b_k z^k +\cdots b_0$, a polynomial of degree $k+1$.  Then by the fundamental theorem of algebra there exists $z_{k+1}$ such that $p(z_{k+1}) = 0$.  Divide $p(z)$ by $(z-z_{k+1})$.  The division rule gives,
\begin{aligned}
p(z) = (z-z_{k+1})q(z) + \gamma 
\end{aligned}
where the degree of $q(z)$ is $k$, being one less than the degree of $p(z)$.  But $p(z_{k+1}) = 0$, and substituting $z_{k+1}$ into this equation shows that $\gamma=0$.  Thus
$$p(z) = (z-z_{k+1})q(z).$$
We can now apply $\mathscr S_k$ to $q(z)$ to find $z_1, \cdots z_k$ such that $$q(z)=(z-z_1)(z-z_2)\cdots(z-z_k),$$ whence
$$p(z) = (z-z_1)(z-z_2)\cdots(z-z_{k+1})
$$
and we have proved $\mathscr S_{k+1}$.  It follows by the principle of induction that $\mathscr S_k$ is true for all $ k \geqslant 1$.
In your proof, you failed to spot that $\gamma=0$, so then couldn't easily proceed.  I also note that the four formulae at each step are not written as equations, which leaves their meaning unclear.
