The theorem which I am referring to states: for any $f, g$ there exist $q, r$ such that $f(x)=g(x)q(x)+r(x)$ with the degree of $r$ less than the degree of $g$ if $g$ is monic.
The book I am using remarks that it can be proven via induction on the degree of $g$, but leaves the proof to the reader. Unfortunately, this reader is not clever enough to get it.
The base case is fairly clear, but I'm completely stuck after that. Any hints?
Hint $ $ If $\, f- g\,h\, $ has $\,\deg \ge g\,$ then subtracting $\, \color{#c00}{c\:\!x^k}\:\!g\ $ (choose $\,\color{#c00}c\,$ to kill lead coef of $\:\!f)$ yields a smaller degree polynomial of same form $\, f - g\, h'.\, $ Note: this inducts on $\,\deg f,\,$ not on $\,\deg g$.
The key idea of polynomial division is this: if the divisor has invertible lead coef $\,b\,$ (e.g. $\,b=1)\,$ and the dividend has degree $\ge$ the divisor, then we can $\rm\color{#c00}{scale}$ the divisor so that it has the same degree and leading coef as the dividend, then subtract it from the dividend, thereby killing the leading term of the dividend; then recursively apply this process to the remaining part of the dividend, which has smaller degree (since we killed the lead term of the dividend), viz.
$$\begin{align} (\overbrace{a x^{\large k+n} + f}^{\large \rm dividend})\ -\ &\color{#c00}{\frac{a}b x^{\large k}} (\overbrace{b x^{\large n} + g}^{\large \rm divisor})\ =\ \overbrace{\color{#0a0}{f-\frac{a}b x^kg}}^{\large {\rm deg}\ <\ k+n}\\[.4em] \Longrightarrow\ \ \ \dfrac{a x^{\large k+n}+f}{bx^{\large n}+g}\, =\ &\color{#c00}{\frac{a}b x^{\large k}}\ \ +\ \underbrace{\dfrac{\color{#0a0}{f-\frac{a}bx^{\large k} g}}{bx^n + g}}_{\large\rm recurse\ on\ this}\end{align}\qquad\qquad$$
$$\require{enclose} \begin{array}{r} \color{#c00}{\frac{a}b\:\! x^k}\phantom{x^{k+n}+f\ }\,\ \ \\[0pt] {\Large \smash[t]{\overset{\rm\color{#90f}{tabularly}}{\leadsto^{\phantom{.}}}}}\quad\ bx^n\!+g\,\ \enclose{longdiv}{\,\ a x^{k+n}\,+\,f\phantom{x^kg}} \\[-3pt] \underline{ax^{k+n}\, +\, \frac{a}b x^k g} \\ \color{#0a0}{f\,-\,\frac{a}b x^k g} \end{array}\qquad\qquad\qquad\qquad\quad$$
where the second equation arises from the first by dividing through by $\,bx^n + g. \,$ The final expression displayed above shows how this single division step is represented in the common $\rm\color{#90f}{table\ form}.\,$ This single division (descent) step is iterated till we reach a dividend that has smaller degree than the divisor (which must occur since $\Bbb N$ is well-ordered; equivalently, we can use a proof by strong induction).
Rephrased in the language of Gaussian elimination: $\,b\,$ being invertible in the lead monomial $\,b\:\!x^n\,$ of the divisor means we can use it as a pivot to scale-up and eliminate all higher degree monomials. This pivoting viewpoint will become clearer when one studies multivariate division algorithms (standard (Grobner) ideal bases and monomial orderings) which are nonlinear generalizations of Gaussian elimination, and unification algorithms for (equational) term rewriting systems.
Remark $ $ This method works $\rm\color{#0a0}{universally}$ to kill the lead coef of the dividend when the lead coef of the divisor $\,g\,$ is a unit (invertible) so it is a divisor of $\rm\color{#0a0}{every}$ coefficient, so it is always possible to scale it to agree with the lead coef of the divisor. Otoh, if the lead coef of the divisor is not a unit then this is not always possible, e.g. consider $\, x \div 2x\:$ in $\rm\,\mathbb Z[x]\!:\: $ if $\,x = (2x)q(x) + c\,$ then evaluating at $\,x=0\Rightarrow c=0,\,$ therefore cancelling $\,x\,$ yields that $\,1 = 2q(x)\,$ thus, again, evaluating at $\,x = 0\Rightarrow 1 = 2q(0)\Rightarrow 2\mid 1\,$ in $\Bbb Z,\,$ contradiction [cf. this proof that $\,x\nmid 1\,$ in $R[x]$].
We can generalize the polynomial division algorithm to divisors with noninvertible lead coefs, e.g.
Theorem (nonmonic Polynomial Division Algorithm) $\ $ Let $\,0\neq F,G\in A[x]\,$ be polynomials over a comm. ring $A,$ with $\,a\,$ = lead coef of $\,F,\,$ and $\, i \ge \max\{0,\,1+\deg G-\deg F\}.\,$ Then
$\qquad\qquad \phantom{1^{1^{1^{1^{1^{1}}}}}}a^{i} G\, =\, Q F + R\ \ {\rm for\ some}\ \ Q,R\in A[x],\ \deg R < \deg F$
Proof $\ $ See here for a few proofs.
There are also multivariate generalizations of the polynomial division algorithm such as the Gröbner basis algorithm. One gains further insight from this more general perspective on the descent process, e.g. in terms of monomial orderings.
HINT:
Prove that in an integral domain, if $f$ and $g$ are nonzero polynomials then $\deg(fg) = \deg(f) + \deg(g)$. Then, once you have the base case and are working with the induction hypothesis, write out the polynomials. That is, $f = a_n x^n + a_{n-2} x^{n-1} + \cdots + a_0$, $g = b_m x^m + \cdots + b_0$. Multiply $g$ by an appropriate multiple of a power of $x$ and subtract. Use the induction hypothesis.
Does that make sense?
Here is a relatively tricky proof from linear algebra done right (but actually not tricky at all). We are aiming to express a polynomial of degree n by two other polynomials with lower degree, n-m for p and m-1 for q seperately, so why not consider the linear spaces Pn-m(F)*Pm-1(F) and Pn(F). T(p,q)=pg+q is obviously a linear map between the two spaces, and it's trivial to check that it's an isomorphism.
