$3^2+2=11$, $33^2+22=1111$, $333^2+222=111111$, and so on. $3^2+2=11$
$33^2+22=1111$
$333^2+222=111111$
$3333^2+2222=11111111$
$\vdots$
The pattern here is obvious, but I could not have a proof.

Prove that $\underset{n\text{ }{3}\text{'s}}{\underbrace{333\dots3}}^2+\underset{n\text{ }{2}\text{'s}}{\underbrace{222\dots2}}=\underset{2n\text{ }{1}\text{'s}}{\underbrace{111\dots1}}$ for any natural number $n$.

Dear, I am not asking you  to prove, I just want a hint, how to start proving it. Thanks.
 A: Because
$$\left(\frac{10^n-1}{3}\right)^2+2\cdot\frac{10^n-1}{9}=\frac{10^{2n}-2\cdot10^n+1+2\cdot10^n-2}{9}=\frac{10^{2n}-1}{9}.$$
A: Hint:
$$\underbrace{aaa\cdots a}_{n\text{ a's}}=\frac{a(10^n-1)}{9}$$
Where $a\in\{0,1,2,3,4,5,6,7,8,9\}$ and $n\in\mathbb{N_0}$.
A: There's also a simple argument using only divisors.
small example:
$$
111111 - 222 = 111 \times 1001 - 111 \times 2 = 111 \times 999 = 333^2
$$
generalized:
$$
\underbrace{11\cdots1}_{2n} - \underbrace{22\cdots 2}_{n} = \underbrace{11\cdot 1}_{n} \times \underbrace{100\cdots01}_{n+1} - 2 \times \underbrace{11\cdots1}_{n} = \underbrace{99\cdots 9}_{n} \times \underbrace{11 \cdots 1}_{n} = (\underbrace{33\cdots3}_{n})^2
$$
A: $${\underbrace{33...3}_{n}}^2+\underbrace{22...2}_{n}=\underbrace{33..3}_{n}\cdot \underbrace{11...1}_{n}\cdot 3+\underbrace{11...1}_{n}\cdot 2=\\
\underbrace{11...1}_{n}(\underbrace{99...9}_{n}+1+1)=\underbrace{11...1}_{n}\cdot 1\underbrace{00...0}_{n}+\underbrace{11...1}_{n}=\underbrace{11...1}_{n}\underbrace{00...0}_{n}+\underbrace{11...1}_{n}=\underbrace{11...1}_{2n}.$$
