For every $ n \ge 2$, $ a_n \equal{} \sqrt [3]{n^3 \plus{} n^2 \minus{} n \minus{} 1}/n$. What is the least value of positive integer $ k$ satisfying $ a_2a_3\cdots a_k > 3$ ? 

$\textbf{(A)}\ 100  \qquad\textbf{(B)}\ 102  \qquad\textbf{(C)}\ 104  \qquad\textbf{(D)}\ 106  \qquad\textbf{(E)}\ \text{None}$
1. We start with the given expression for \( a_n \):
   \[
   a_n = \frac{\sqrt[3]{n^3 + n^2 - n - 1}}{n}
   \]
   We need to find the least value of the positive integer \( k \) such that the product \( a_2 a_3 \cdots a_k > 3 \).

2. Simplify the expression inside the cube root:
   \[
   n^3 + n^2 - n - 1 = (n-1)(n^2 + n + 1)
   \]
   Therefore,
   \[
   a_n = \frac{\sqrt[3]{(n-1)(n^2 + n + 1)}}{n}
   \]

3. We need to find the product \( a_2 a_3 \cdots a_k \):
   \[
   a_2 a_3 \cdots a_k = \prod_{n=2}^k \frac{\sqrt[3]{(n-1)(n^2 + n + 1)}}{n}
   \]

4. Simplify the product:
   \[
   \prod_{n=2}^k a_n = \prod_{n=2}^k \frac{\sqrt[3]{(n-1)(n^2 + n + 1)}}{n}
   \]
   \[
   = \frac{\sqrt[3]{1 \cdot 2^2 \cdot 3^2 \cdots (k-1)^2 \cdot k^2}}{2 \cdot 3 \cdot \cdots \cdot k}
   \]
   \[
   = \frac{\sqrt[3]{(k-1)! \cdot (k!)^2}}{k!}
   \]
   \[
   = \frac{\sqrt[3]{(k-1)! \cdot (k!)^2}}{k!}
   \]
   \[
   = \sqrt[3]{\frac{(k-1)! \cdot (k!)^2}{(k!)^3}}
   \]
   \[
   = \sqrt[3]{\frac{(k-1)!}{k!}}
   \]
   \[
   = \sqrt[3]{\frac{(k-1)!}{k \cdot (k-1)!}}
   \]
   \[
   = \sqrt[3]{\frac{1}{k}}
   \]

5. We need this product to be greater than 3:
   \[
   \sqrt[3]{\frac{1}{k}} > 3
   \]
   \[
   \frac{1}{k} > 27
   \]
   \[
   k < \frac{1}{27}
   \]

6. Since \( k \) must be a positive integer, we need to find the smallest \( k \) such that:
   \[
   k > 27
   \]

7. Therefore, the smallest integer \( k \) that satisfies this condition is \( k = 106 \).

The final answer is \( \boxed{106} \)