Question 33
Suppose
$$\frac1{2013^{1000}}=a_1+\frac{a_2}{2!}+\frac{a_3}{3!}+\cdots+\frac{a_n}{n!},$$
where $n$ is a positive integer, $a_1,\ldots,a_n$ are nonnegative integers such that $a_k\lt k$ for $k=2,\ldots,n$ and $a_n>0$. Find $n$.
$$\frac1{2013^{1000}}=a_1+\frac{a_2}{2!}+\frac{a_3}{3!}+\cdots+\frac{a_n}{n!},$$
where $n$ is a positive integer, $a_1,\ldots,a_n$ are nonnegative integers such that $a_k\lt k$ for $k=2,\ldots,n$ and $a_n>0$. Find $n$.