Let $a_{1}, a_{2}, \ldots, a_{1000}$ be real numbers such that

$$
\begin{array}{r}
a_{1} \cdot 1+a_{2} \cdot 2+a_{3} \cdot 3+\cdots+a_{1000} \cdot 1000=0 \\
a_{1} \cdot 1^{2}+a_{2} \cdot 2^{2}+a_{3} \cdot 3^{2}+\cdots+a_{1000} \cdot 1000^{2}=0 \\
a_{1} \cdot 1^{3}+a_{2} \cdot 2^{3}+a_{3} \cdot 3^{3}+\cdots+a_{1000} \cdot 1000^{3}=0 \\
\vdots \\
a_{1} \cdot 1^{999}+a_{2} \cdot 2^{999}+a_{3} \cdot 3^{999}+\cdots+a_{1000} \cdot 1000^{999}=0 \\
a_{1} \cdot 1^{1000}+a_{2} \cdot 2^{1000}+a_{3} \cdot 3^{1000}+\cdots+a_{1000} \cdot 1000^{1000}=1 .
\end{array}
$$

What is the value of $a_{1}$ ?
The key idea is to multiply on weights which are polynomial coefficients. Specifically, let $P(n)=n(n-2)(n-3) \ldots(n-1000)=\sum_{k=0}^{1000} c_{k} n^{k}$, say. Note that $c_{1}=0$ and $c_{1000}=1$.

Now take the $k$ th equation and multiply it by $c_{k}$, then add all equations together. From this we obtain

$$
a_{1} P(1)+a_{2} P(2)+\cdots+a_{1000} P(1000)=c_{1000}=1 .
$$

But $P(1)=(-1)^{999} \cdot 999$ ! and $P(2)=\cdots=P(1000)=0$. Therefore, $a_{1}=\frac{-1}{999 !}$.