I am trying to find if
$$\prod_{j=0}^{k}(1-a_{k-j})$$
can be lower bounded by some constant $c$, with $0 \leq a_n \leq 1$. I have seen in other post that
$$\prod \limits_{j=0}^{k}(1-a_{k-j}) \geq1- \sum\limits_{j=k}^{n}a_{k-j}$$
but I do not know how to proceed. Given $\lim_{m\to \infty} a_m \to 0$ can we find such a lower bound $c$?
We cannot bound this product with some $c>0$. For example, let us take a sequence that satisfy the given conditions: $$a_n = \begin{cases} 2, & n = 0 \\ 1/2, & n = \overline{1,k} \\ 0, & n > k \end{cases}$$ Definitely, $a_n \to 0$ and $a_n \ge 0$. However, $$\prod_{j=0}^{k}(1-a_{k-j}) = \prod_{j=0}^k(1-a_j) = (1-2)\left(1 - \frac{1}{2}\right)^k < 0$$ Note that by choosing $a_0$ appropriately, we can make this product be less than any real number.
Therefore, with no constraints on $a_n$, we cannot lower bound the product.
EDIT: After the additional condition $0 \le a_n \le 1$, we have $$1-a_0 \ge 1-1 = 0 \\ 1-a_1 \ge 1-1 = 0 \\ ... \\ 1-a_k \ge 1-1 = 0 $$ Multiplying all inequalities we get $$\prod_{j=0}^k(1-a_j) \ge 0$$