$3^2+2=11$, $33^2+22=1111$, $333^2+222=111111$, 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.

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 $n\in\mathbb{N_0}$.
