Convergence of a product of polynomials Let $m$ be a positive integer, and define $f_n\colon[0,1]\to[0,1]$ recursively as follows:
$$f_0(x)=(1-x)^m.$$
$$f_{n+1}(x)=f_n(x)\cdot(1-x^k)\qquad\text{if }f_n(x)-1+mx\in\Theta(x^k).$$
I am pretty sure that this converges to $f(x)=\max\{1-mx,0\}$. In fact, if we write
$$
f_n(x)=1-mx+ax^k+O(x^{k+1})\text{ with }a\ne0,
$$
then $f_{n+1}(x)=1-mx+(a-1)x^k+O(x^{k+1})$ so that the coefficient $a$ gets consumed step by step until it is zero and then the next power is attacked. However, this would require that the integer $a$ is never negative. According to numerical experiments, this indeed seems to be the case, but I don't see a compelling argument for this.
I tried to show that $f_n\ge f$ or that $f_n(\frac1m)\to 0$, but ended up running in circles.
Q: Can anyone prove (or disprove) that $f_n\to f$ and in particular that the first non-zero coefficient $a$ above is always positive?

Here are the first terms of the sequence for $m=2$:
$$\begin{align}
f_0(x)=(1-x)^2&=1-2x+x^{\color{red}2}\\
f_1(x)=f_0(x)(1-x^{\color{red}2})&=1-2x+2x^{\color{red}3}-x^4\\
f_2(x)=f_1(x)(1-x^{\color{red}3})&=1-2x+x^{\color{red}3}+x^4-2x^6+x^7\\
f_3(x)=f_2(x)(1-x^{\color{red}3})&=1-2x+3x^{\color{red}4}-3x^6+2x^9-x^{10}\\
f_4(x)=f_3(x)(1-x^{\color{red}4})&=1-2x+2x^{\color{red}4}+2x^5-3x^6+\cdots\\
f_5(x)=f_4(x)(1-x^{\color{red}4})&=\cdots
\end{align}$$
And here's a plot of
$$\begin{align}f_{69}(x)=&(1-x)^2(1-x^2)(1-x^3)^2\cdot\\& (1-x^4)^3(1-x^5)^6(1-x^6)^9\cdot\\&(1-x^7)^{18}(1-x^8)^{30}\end{align} $$

(So apparently, for $m=2$, there is a relation to A001037 or A059966 or A060477)
 A: Apparently, I stumbled over the cyclotomic identity
$${1 \over 1-\alpha z}=\prod_{j=1}^\infty\left({1 \over 1-z^j}\right)^{M(\alpha,j)} $$ of (formal) power series
where the detail that matters is that $M(\alpha,j)$ is a non-negative integer because it counts something.
Taking reciprocals and renaming variables as in my question, we get
$$\tag2 1-dx=(1-x)^d\cdot \prod_{k=2}^\infty(1-x^k)^{a_k}$$
where the $a_k$ are non-negative integers. While this equality seems absurd when $\frac1d<x\le 1$, when the left hand side is negative and the right hand side is a product of non-negative numbers, keep in mind that an infinite product does not converge when the partial product of all non-zero factors converges to $0$ or when infinitely many of the factors are $=0$. Hence $(2)$ makes a statement about the sequence of partial products only for $0\le x<\frac1d$, whereas the partial product then necessarily converges point-wise to $0$ for $\frac1d\le x\le 1$.
If we consider such a partial product, then none of the later factors can change the low order terms of the power series so that
$$ (1-x)^d\prod_{k=1}^m(a-x^k)^{a_k}=1-dx+O(x^{m+1}). $$
This also means that
$$ (1-x)^d\prod_{k=1}^{m-1}(a-x^k)^{a_k}=1-dx+a_mx^m+O(x^{m+1}).$$
Going backwards, we conclude that the recursion defined in the OP will indeed use multiplication by $(1-x^k)$ in exactly $a_k\in\Bbb N_0$ steps, and also that $f_n\to f$ as conjectured.
