



 
Jash, The proof is not hard to find; consider the smallest number with $n$ digits $m^{n1}$, and the largest $m^n  1.$ Show that the quotient of their squares \[\frac{(m^n  1)^2}{(m^{n1})^2}\] is always strictly bigger than $m.$ Deduce that $(m^n  1)^2$ always has exactly one more digit than $(m^{n1})^{2}.$ Good Hunting!  


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