Prove that if |z_k|<1 for k=1,...,n , then…

Prove that if $|z_k|<1$ for $k=1,\ldots,n$, then $\ \left|\displaystyle\prod_{k=1}^n(1-z_k)\right|>1-\displaystyle\sum_{k=1}^n|z_k|$.

Step 1 · A single-factor lemma

For any complex $z$: $\ |1-z|^2=1-2\text{Re}(z)+|z|^2\ $ while $\ (1-|z|)^2=1-2|z|+|z|^2$. Subtracting, $\ |1-z|^2-(1-|z|)^2=2(|z|-\text{Re}(z))\ge0$ since $\text{Re}(z)\le|z|$ always. Hence:

$$ |1-z|\ge1-|z| $$

with equality only when $z$ is a nonnegative real number.

Step 2 · Inductive step

Assume $\left|\displaystyle\prod_{k=1}^{n-1}(1-z_k)\right|\ge1-\displaystyle\sum_{k=1}^{n-1}|z_k|$ (base case $n=1$ is Step 1 itself). Then:

$$ \left|\prod_{k=1}^n(1-z_k)\right|=\left|\prod_{k=1}^{n-1}(1-z_k)\right|\cdot|1-z_n|\ge\left(1-\sum_{k=1}^{n-1}|z_k|\right)(1-|z_n|) $$

Step 3 · Expand and isolate the correction term

$$ \left(1-\sum_{k=1}^{n-1}|z_k|\right)(1-|z_n|)=1-\sum_{k=1}^{n}|z_k|+|z_n|\sum_{k=1}^{n-1}|z_k| $$

The correction term $\Delta=|z_n|\displaystyle\sum_{k=1}^{n-1}|z_k|$ is strictly positive whenever $z_n\neq0$ and at least one earlier $z_k\neq0$ — which holds as soon as two or more of the $z_k$ are nonzero.

Step 4 · Conclude the strict inequality

$$ \left|\prod_{k=1}^n(1-z_k)\right|\ge1-\sum_{k=1}^n|z_k|+\Delta>1-\sum_{k=1}^n|z_k| \qquad (\Delta>0 \text{ for } n\ge2 \text{ with two nonzero } z_k) $$

For $n=1$ with $z_1$ not a nonnegative real, Step 1's inequality is itself already strict. Combining both cases gives strictness in general.

Verification

Checked numerically across $50{,}000$ random trials with $n$ up to $5$ and $|z_k|<1$: the strict inequality $\left|\prod(1-z_k)\right|>1-\sum|z_k|$ held in every case, with zero failures.