In this paper by Jerry Hu, he defines the function
$$f_{s,k,i}\left(u\right)=\prod_{p\mid u} \left(1-\frac{\sum_{m=i}^{k-1}{s \choose m}\left(p-1\right)^{k-1-m}}{\sum_{m=0}^{k-1}{s \choose m}\left(p-1\right)^{k-1-m}}\right)$$
for use in the main theorem.
In Lemma 4 (pages 3 & 4), it is claimed that
$$\frac{f_{s,k,i}\left(u_{i}\right)}{f_{s,k,i+1}\left(u_{i}\right)}=\sum_{d\mid u_{i}}\frac{\mu\left(d\right){s \choose i}^{\omega\left(d\right)}}{\alpha_{s,k,i}\left(d\right)}$$
for $i=1,2,\ldots k-2$, where $\mu$ is the Möbius function, $\omega\left(d\right)$ is the number of distinct prime factors of $d$, and
$$\alpha_{s,k,i}\left(d\right)=d^{i}\prod_{p\mid d}\sum_{m=0}^{i}{s \choose m}\left(1-p^{-1}\right)^{i-m}p^{-m}$$
for $i=1,2,\ldots k-1$.
Obviously $f_{s,k,i}$ is multiplicative and, hence, so is $\frac{f_{s,k,i}}{f_{s,k,i+1}}$, which means that it is enough to show this for $u_{i}=p^{a}$ for integers $a\geq0$.
I've only been able to go so far on either side of the equality, but I've hit a wall.
Starting from the left hand side, since the only prime dividing $p^{a}$ is $p$ itself, I have:
$$\frac{f_{s,k,i}\left(p^{a}\right)}{f_{s,k,i+1}\left(p^{a}\right)}=\frac{1-\frac{\sum_{m=i}^{k-1}{s \choose m}\left(p-1\right)^{k-1-m}}{\sum_{m=0}^{k-1}{s \choose m}\left(p-1\right)^{k-1-m}}}{1-\frac{\sum_{m=i+1}^{k-1}{s \choose m}\left(p-1\right)^{k-1-m}}{\sum_{m=0}^{k-1}{s \choose m}\left(p-1\right)^{k-1-m}}}= \\ \frac{\frac{\sum_{m=0}^{i-1}{s \choose m}\left(p-1\right)^{k-1-m}}{\sum_{m=0}^{k-1}{s \choose m}\left(p-1\right)^{k-1-m}}}{\frac{\sum_{m=0}^{i}{s \choose m}\left(p-1\right)^{k-1-m}}{\sum_{m=0}^{k-1}{s \choose m}\left(p-1\right)^{k-1-m}}} = \frac{\sum_{m=0}^{i-1}{s \choose m}\left(p-1\right)^{k-1-m}}{\sum_{m=0}^{i}{s \choose m}\left(p-1\right)^{k-1-m}}.$$
On the other hand, since $\mu\left(1\right)=1$, $\mu\left(p\right)=-1$, and $\mu\left(p^{a}\right)=0$ for $a>1$, as well as since $\omega\left(1\right)=0$, $\omega\left(p^{a}\right)=1$ for $a\geq1$, and $\alpha_{s,k,i}\left(1\right)=1$, I have:
$$\sum_{d\mid p^{a}}\frac{\mu\left(d\right){s \choose i}^{\omega\left(d\right)}}{\alpha_{s,k,i}\left(d\right)}=1-{s \choose i}\left(\alpha_{s,k,i}\left(p\right)\right)^{-1} = \\ 1-{s \choose i}\left(p^{i}\sum_{m=0}^{i}{s \choose m}\left(1-p^{-1}\right)^{i-m}p^{-m}\right)^{-1} = \\ 1-{s \choose i}\left(p^{i}\sum_{m=0}^{i}{s \choose m}\left(p-1\right)^{i-m}p^{-i}\right)^{-1} = \\ 1-{s \choose i}\left(\sum_{m=0}^{i}{s \choose m}\left(p-1\right)^{i-m}\right)^{-1}.$$
I can't see how to proceed from here. Assuming I haven't made any errors, what should I do to fill in the gap in
$$\frac{\sum_{m=0}^{i-1}{s \choose m}\left(p-1\right)^{k-1-m}}{\sum_{m=0}^{i}{s \choose m}\left(p-1\right)^{k-1-m}}=\overset{?}{\cdots}=1-{s \choose i}\left(\sum_{m=0}^{i}{s \choose m}\left(p-1\right)^{i-m}\right)^{-1}?$$
The second part of this lemma looks similar, so I think I can do that one once I've understood how to move forward in this one.
EDIT: I somehow missed an important part of the definition of $f_{s,k,i}$ when typing this question, and this continued throughout due to my copying and pasting. This has been fixed. I'm sorry for not having noticed this earlier.
Here is one way to do the calculation:
\begin{align*} \frac{\sum_{m=0}^{i-1}\binom{s}{m}(p-1)^{k-1-m}}{\sum_{m=0}^{i}\binom{s}{m}(p-1)^{k-1-m}}&= \frac{\sum_{m=0}^{i-1}\binom{s}{m}(p-1)^{-m}}{\sum_{m=0}^{i}\binom{s}{m}(p-1)^{-m}}\tag{1}\\ &=\frac{\sum_{m=0}^{i}\binom{s}{m}(p-1)^{-m}-\binom{s}{i}(p-1)^{-i}}{\sum_{m=0}^{i}\binom{s}{m}(p-1)^{-m}}\tag{2}\\ &=1-\binom{s}{i}(p-1)^{-i}\left(\sum_{m=0}^{i}\binom{s}{m}(p-1)^{-m}\right)^{-1}\tag{3}\\ &=1-\binom{s}{i}\left(\sum_{m=0}^{i}\binom{s}{m}(p-1)^{i-m}\right)^{-1}\tag{4}\\ \end{align*}
Comment:
In (1) we divide by $(p-1)^{k-1}$
In (2) we add the summand $m=i$ to the sum in the numerator and subtract the corresponding value
In (3) we perform the division
In (4) there is a last small rearrangement
Note: According to your precise description you did most of the analysis of Lemma 4 and I suppose that this calculation will look therefore pretty easy to you.