Please help me to solve my homework ;)
Prove that for any positive integer $n$ a square of rather big number divides even more huge number:
$${\LARGE \left.\underbrace{33\dots 3}_{1\underbrace{00\dots 0}_n}\underbrace{66\dots 6}_{\underbrace{99\dots 9}_n}7^2\right\vert
1\underbrace{00\dots 0}_{\underbrace{99\dots 9}_{{1\underbrace{00\dots 0}_n}+n }}1\underbrace{00\dots 0}_{\underbrace{99\dots 9}_n}1}$$
Blocks of repeated digits are specified by their lengths, sometimes in a nested manner.
EDIT :
Some examples (according to Bill's request) for $n=0,1,2$:
$$\LARGE 37^2\Biggm| 100000000011$$
$$\LARGE 33333333336666666667^2\Biggm| 1\underbrace{00\dots0}_{10^{11}-1}10000000001$$
$$\LARGE \underbrace{33\dots3}_{100}\underbrace{66\dots6}_{99}7^2\Biggm| 1\underbrace{00\dots0}_{10^{102}-1}1\underbrace{00\dots0}_{99}1$$
Hint: First, write the numbers as proper algebraic expressions. Note for example that $\underbrace{11\ldots1}_n=\frac{10^n-1}{9}$. That makes the (not yet squared) number on the left hand side $$ \underbrace{33\ldots3}_{10^n}\underbrace{66\ldots 6}_{10^n-1}7=\underbrace{33\ldots3}_{10^{2n}}+\underbrace{33\ldots3}_{10^{n}}+1=\frac{10^{10^{2n}}-1}{3}+\frac{10^{10^{n}}-1}{3}+1.$$ Do a similar rewrite on the right hand side and make use of several standard facts such as $a^n-1\mid a^m-1$ if $n\mid m.