inequality proof consisting of complex number modulus and intergers 
Let $m,n$ be integers greater than $1$.  Let $\zeta$ be a complex number. Suppose $0<|\zeta|<\frac12$. Then
$$
\frac{1-|\zeta|-|\zeta|^m}{1-|\zeta|}<|(1+\zeta^m)(1+\zeta^{m+1})\cdot …\cdot (1+\zeta^{m+n})|<\frac{1-|\zeta|}{1-|\zeta|-|\zeta|^m}.
$$

I tried to use induction to prove.
$P(n)$:
$\dfrac{1-|\zeta|-|\zeta|^m}{1-|\zeta|}<|(1+\zeta^m)(1+\zeta^{m+1})\cdot …\cdot (1+\zeta^{m+n})|$
$P(2)$: (I tried to “simplify” the inequality)
$\dfrac{1-|\zeta|-|\zeta|^m}{1-|\zeta|}<|(1+\zeta^m)(1+\zeta^{m+1})(1+\zeta^{m+2})|$
$1-\dfrac{|\zeta|^m}{1-|\zeta|}<|(1+\zeta^m)(1+\zeta^{m+1})(1+\zeta^{m+2})|$
$1-|\zeta|^m(\sum_{x=0}^{\infty}|\zeta|^x)<|(1+\zeta^m)(1+\zeta^{m+1})(1+\zeta^{m+2})|$
Then I stuck here, can anyone give me some hints?
Edit:
I have another thought just now.
$\frac{1-|\zeta|-|\zeta|^m}{1-|\zeta|}<|(1+\zeta^m)(1+\zeta^{m+1})\cdot …\cdot (1+\zeta^{m+n})|$
$1-\frac{|\zeta|^m}{1-|\zeta|}<|(1+\zeta^m)(1+\zeta^{m+1})\cdot …\cdot (1+\zeta^{m+n})|$
$1<|(1+\zeta^m)(1+\zeta^{m+1})\cdot …\cdot (1+\zeta^{m+n})|+ \frac{|\zeta|^m}{1-|\zeta|}$
Now, let $P(n)$ be $1<|(1+\zeta^m)(1+\zeta^{m+1})\cdot …\cdot (1+\zeta^{m+n})|+ \frac{|\zeta|^m}{1-|\zeta|}$
This time, I start at $P(0)$:
$|1+\zeta^m|+\frac{|\zeta|^m}{1-|\zeta|}>|1+\zeta^m|+|\zeta|^m= |1+\zeta^m|+|\zeta^m|\ge |1+2\zeta^m|$
Then I stuck here.
Second Edit:
I have proved that:
(A):
$\frac{1}{1-|\zeta|}\ge 1+|\zeta|\ge |1+\zeta|\ge 1-|\zeta|$
(B):
$|(1+\zeta^m)\cdot (1+\zeta^{m+1})\cdot …\cdot (1+\zeta^{m+n})$
$=|1+\zeta^m|\cdot |1+\zeta^{m+1}|\cdot …\cdot |1+\zeta^{m+n}|$
$\ge (1-|\zeta|^m)\cdot (1-|\zeta|^{m+1})\cdot …\cdot (1-|\zeta|^{m+n})$
(C):
$1-|\zeta|^m-|\zeta|^{m+1}-…-|\zeta|^{m+n}$
$\le (1-|\zeta|^m)\cdot (1-|\zeta|^{m+1})\cdot …\cdot (1-|\zeta|^{m+n})$
$\le \frac{1}{1+|\zeta|^m+|\zeta|^{m+1}+…+|\zeta|^{m+n}}$
Then I stuck here.
Third edit:
(C) (rectified):
$1-|z|^m-\frac{|z|^{m+1}(1-|z|^n)}{1-|z|}$
$1-|z|^m-|z|^{m+1}-…-|z|^{m+n}$
$\le (1-|z|^m)\cdot (1-|z|^{m+1})\cdot …\cdot (1-|z|^{m+n})$
$\le \frac{1}{1+|z|^m+|z|^{m+1}+…+|z|^{m+n}}=\frac{1}{1+|z|^m+ \frac{|z|^{m+1}(1-|z|^n)}{1-|z|}}$
I am not sure whether this is useful so I don’t combine this into the second edit.
 A: For the right inequality:
Let $x = |\zeta |$. We have
\begin{align*}
 \frac{1 - x}{1 - x - x^m} &= \frac{1}{1 - x^m/(1 - x)} \\[5pt]
 &> \mathrm{exp}\left({x^m/(1 - x)}\right)\tag{1}\\
 &= \mathrm{exp}\left({\sum_{k=m}^\infty x^k}\right) \\
 &\ge \mathrm{exp}\left({\sum_{k=m}^\infty \ln(1 + x^k)}\right)\tag{2}\\
 &= \prod_{k=m}^\infty (1 + x^k)\\
 &\ge \prod_{k=m}^{m+n} (1 + x^k)
\end{align*}
where we have used $\mathrm{e}^{-v} > 1 - v$ for all $0 < v < 1$ in (1), and $\ln(1 + u) \le u$ for all $u \ge 0$ in (2).
Using $|1 + \zeta^k| \le 1 + |\zeta|^k$, the right inequality is true.

For the left inequality:
Fact 1: Let $y_k \in [0, 1), \, \forall k$. Then $(1 - y_1)(1 - y_2)\cdots (1-y_N)
\ge 1 - (y_1 + y_2 + \cdots + y_N)$.
(Note: Use Mathematical Induction and $(1 - a)(1 - b) \ge 1 - (a + b)$ for all $a, b \ge 0$.)
Using Fact 1, we have
\begin{align*}
 \frac{1 - x - x^m}{1 - x} &= 1 - x^m/(1 - x)\\
 &= 1 - \sum_{k=m}^\infty x^k\\ 
 &< 1 - \sum_{k=m}^{m+n} x^k\\
 &\le \prod_{k=m}^{m+n} (1 - x^k).
\end{align*}
Using $|1 + \zeta^k| \ge 1 - |\zeta|^k$, the left inequality is true.
