We only need to prove that there are positive integers $n$, $k$ such that
$$\underbrace{11 \dots 11}_{1992} \cdot 10^k \le n^{1992} < \underbrace{11 \dots 1}_{1991} 2 \cdot 10^k \\ \iff \sqrt[1992]{11 \dots 11} \cdot 10^t \le n < \sqrt[1992]{11 \dots 12} \cdot 10^t$$
in which $k = 1992t$.
Suppose the decimal expansions of $\sqrt[1992]{11 \dots 11}$ and $\sqrt[1992]{11 \dots 12}$ differ in the $t$th position. So
$$n = \left\lceil \sqrt[1992]{\underbrace{11 \dots 11}_{1992}} \cdot 10^t \right\rceil + 1$$
suffices.