TITLE: Prove that there is an integer $n$ such that $n^{1992}$ starts with $1992$ one's.
QUESTION [8 upvotes]: This was taken from an old Brazilian Mathematical Olympiad (1992).
As the title says, we're supposed to prove that there is an integer $n$ such that $n^{1992}$ starts with $1992$ one's (in the decimal expansion).
In principle I'm expecting a proof using rather elementary methods, though any solution at all is appreciated and will be upvoted.

REPLY [3 votes]: Let $a=111...111$ Consider 
$$a\times 10^k\lt n^{1992}\lt (a+1)\times 10^k \tag{*} $$
If $k$ sufficient large, there must be an integer $n$ such that
$$\sqrt[1992]{a}\times  10^{\frac k{1992}}\lt   n\lt\sqrt[1992]{a+1}\times  10^{\frac k{1992}} $$
that is $(*)$


In general, If $f(x)$ is a polynomial, $a\in \Bbb N$, there must be $n\in \Bbb N$ such that
$f(n)$ starts with $a$

In fact, If $k$ sufficient large, there must be an integer $n$ such that
$$f(n)\lt a\times10^k\lt f(n+1)\qquad f(n+1)-f(n)\lt 10^k$$