Solution:
We first find a multiple of $5^{1000}$ which has no zeros in the last $1000$ digits. Suppose that we have a multiple $n \cdot 5^{1000}$ whose last zero is in place $r$ (treating the last place as place $0$, the next to last as place $1$ and so on). Then $n(10^{r} + 1)$ has the same digits in places $0$ to $r-1$ and a nonzero digit in place $r$, and hence no zeros in places $0$ to $r$. So repeating, we find a multiple $n \cdot 5^{1000}$ with no zeros in the last $1000$ digits.

Now let $m$ be the remainder when $n$ is divided by $2^{1000}$, so $n = k \cdot 2^{1000} + m$, and hence $m \cdot 5^{1000} = n \cdot 5^{1000} - k \cdot 10^{1000}$. So $m \cdot 5^{1000}$ has the same last $1000$ digits as $n \cdot 5^{1000}$. But it has less than $1001$ digits, and hence it has exactly $1000$ digits and no zeros.