Prove that exists a positive integer $n$ such that first $2016$ digits of $n^{2016}$ are all 1. Prove that exists a positive integer $n$ such that first $2016$ digits of $n^{2016}$ are all 1.
Tip: see the decimal representation of $\sqrt[2016]{\dfrac{1}{9}}$.
I just made this:
$$
\sqrt[2016]{\dfrac{1}{9}} = \sqrt[2016]{0,111\dots} = k
$$
$$
k^{2016} = 0,111\dots
$$
$$
(10k)^{2016} = \underbrace{111\dots111}_{2016 {\ \ }times},111\dots = \underbrace{111\dots111}_{2016{\ \ }times} + \dfrac{1}{9}
$$
$$
10^{p}(10k)^{2016} - \dfrac{10^{p}}{9} = \underbrace{111\dots111}_{2016{\ \ }times}.10^{p}
$$
And thats all. Help me, please, i can't continue from this point.
 A: Consider $a=11\ldots 1 $ with $2016$ ones . 
Take the numbers $11\ldots100\ldots 0=10^k \cdot a$ and $11\ldots120\ldots 0 =10^k \cdot (a+1)$ .
I'll show that between them there is some number of the form $n^{2016}$ for a sufficiently big $k$ which will prove the claim .
For such a power to exist it's sufficiently that : 
$$\sqrt[2016]{10^k \cdot (a+1) }-\sqrt[2016]{10^k \cdot a} > 1$$
Denote these numbers with $x$ and $y$ .
Now use the well-known formula : 
$$x-y=\frac{x^{2016}-y^{2016}}{x^{2015}+x^{2014}y+\ldots+y^{2015}}$$ 
Let's make a trivial bound on the denominator using $y<x$ :
$$x^{2015}+x^{2014}y+\ldots+y^{2015}<x^{2015}+x^{2015}+\ldots+x^{2015}=2016 x^{2015}$$
Also note that $$x^{2016}-y^{2016}=10^k(a+1-a)=10^k$$
Using all these we can achieve the bound : 
$$x-y >\frac{10^k}{2016x^{2015}}$$
We want $x-y > 1$ so let's try to see if $$\frac{10^k}{2016x^{2015}} \geq 1$$
Raise this to the $2016$-th power :
$$10^{2016k} \geq 2016^{2016} \cdot 10^{2015k} \cdot (a+1)^{2015}$$
$$10^k \geq 2016^{2016} \cdot (a+1)^{2015}$$ which is true for a very big $k$ because on the RHS there is a constant number .
This means that there are an infinitely many powers $n^{2016}$ which begin with $2016$ ones .
