8. Let $p_{k}=1+\frac{1}{k}-\frac{1}{k^{2}}-\frac{1}{k^{3}}$, where $k$ is a positive integer. Find the least positive integer $n$ such that the product $p_{2} p_{3} \ldots p_{n}$ exceeds 2010 .
8. Answer: 8038
First, note that
$$
p_{k}=1+\frac{1}{k}-\frac{1}{k^{2}}-\frac{1}{k^{3}}=\left(1-\frac{1}{k}\right)\left(1+\frac{1}{k}\right)^{2}=\frac{(k-1)(k+1)^{2}}{k^{3}} .
$$

Therefore,
$$
p_{2} p_{3} \cdots p_{n}=\frac{1 \cdot 3^{2}}{2^{3}} \cdot \frac{2 \cdot 4^{2}}{3^{3}} \cdot \frac{3 \cdot 5^{2}}{4^{3}} \cdots \frac{(n-1)(n+1)^{2}}{n^{3}}=\frac{(n+1)^{2}}{4 n} .
$$

Next, observe that $\frac{(n+1)^{2}}{4 n}>2010$ is equivalent to $n^{2}-8038 n+1>0$, which is equivalent to $n(n-8038)>-1$. Since $n$ is a positive integer, the last inequality is satisfied if and only if $n \geq 8038$. Consequently, the least $n$ required is 8038 .