Let $\{x_{k}\}$ be a bounded sequence, i.e., $|x_{k}|\le T$ for some positive constant $T$. Also, $x_{k}\neq 1$ and $x_{k}\neq 0$. If \begin{align*} \lim_{t\to \infty}\prod_{k=0}^{t}(1-x_{k})=0 \end{align*} exponentially(at a rate of expoential decaying).
The question is can we find a positive constant $c$ relies only on $T$, such that \begin{align*} \prod_{k=0}^{t}x_{k}\le \alpha|x_{0}|c^{t} \end{align*} for some positive constant $\alpha$？
My initial thoughts:
As pointed out by @C Monsour, we can construct $x_{k}$ to be such that \begin{align} x_{k}=&T, ~~\text{if}~~~ k ~~~\text{is even};\\ x_{k}=&\frac{T-1}{T}, ~~\text{if}~~~ k ~~~\text{is odd}. \end{align} Thus, we have \begin{align} |(1-x_{k})(1-x_{k+1})|=|\frac{1-T}{T}|<1. \end{align} Then, for every even $t$, \begin{align*} \prod_{k=0}^{t}x_{k}=&x_{0}(T\frac{T-1}{T})^{0.5t}\\ =&x_{0}\sqrt{T-1}^{t}\\ \le&\alpha|x_{0}| c^{t}, \end{align*} where $\alpha=1$ and $c=\sqrt{T-1}$.
Can we make the hypothesis that the value of $c$ depends only on the bound of $x_{k}$, i.e., only depends on $T$?