Originally Posted by
newtoinequality For any integer $\displaystyle n>1$, prove that there is exactly one power of 2 having exactly n digits with leading digit 1
First, show existence. If the statement were false, then there would exist $\displaystyle m,\,n$ such that $\displaystyle 2^m<10^{n-1}$ and $\displaystyle 2^{m+1}\ge2\cdot10^{n-1}.$ But the former gives $\displaystyle 2^{m+1}<2\cdot10^{n-1}$ which is a contradiction of the latter.
Hence for each $\displaystyle n$ there must be $\displaystyle m$ such that $\displaystyle 10^{n-1}\le2^m<2\cdot10^{n-1},$ i.e.
$\displaystyle 2^m\ <\ 2\cdot10^{n-1}\quad\ldots\ \fbox1$
$\displaystyle 2^m\ \ge\ 10^{n-1}\quad\ldots\ \fbox2$
To prove uniqueness, observe that $\displaystyle 2^{m-1}<10^{n-1}$ (from $\displaystyle \fbox1)$ and $\displaystyle 2^{m+1}\ge2\cdot10^{n-1}$ (from $\displaystyle \fbox2).$ The result follows since if $\displaystyle m'\ne m$ then either $\displaystyle 2^{m'}\le2^{m-1}$ or $\displaystyle 2^{m'}\ge2^{m+1}$ and so $\displaystyle 2^{m'}$ cannot be between $\displaystyle 10^{n-1}$ and $\displaystyle 2\cdot10^{n-1}.$