Can every odd integer greater than $1$ be written as a product of fractions of the form $\frac{4m+1}{2m+1}$, $m$ a positive integer?

Here is a proof if the fractions were instead $\frac{4m-1}{2m+1}$. I tried to mimic it, but it doesn't work.

---

Since the fractions available have odd numerator and denominator, the same will be true of any product, so even integers aren't reachable. Odd integers can be reached, however.

Let $f(q)$ be the fraction $\frac{4q-1}{2q+1}$. Then $f(1)=1$, $f(4)f(7)=3$ and $3 f(4)=5$. So $1,3,5$ are obtainable.

Suppose now that $m\ge 2$ and we've obtained all odd numbers up to and including $4m-3$. Since $2m+1\le4m-3$ we can use $2m+1$ to obtain
\begin{gather\*}
4m-1 = (2m+1) f(m),\\
4m+1=(2m+1) f(3m+1).
\end{gather\*}
Hence $4m-1$ and $4m+1$ are obtainable as well, so having obtained $5$ we can then reach all odd positive integers.

---

Any help appreciated!