How to prove $(6666\ldots66)^2 + 8888\ldots88 = 4444\ldots44$ (with $n$ 6s and 8s, $2n$ 4s)

Given $$\underbrace{(666666666666\ldots)^2}_{n~\text{times}}+\underbrace{(888888888888\ldots)}_{n~\text{times}}$$

Now we can write $$\underbrace{666666666666\ldots}_{n~\text{times}} = (6+6\cdot 10+6\cdot 10^2+\cdots+6\cdot 10^{n-1})$$

$$\displaystyle =6\left[\frac{10^n-1}{10-1}\right] = \frac{2}{3}\left[10^n-1\right]$$

Similarly we can write $$\underbrace{88888888888\ldots}_{n~\text{times}} = (8+8\cdot 10+8\cdot 10^2+\cdots+8\cdot 10^{n-1})$$

$$\displaystyle =8\left[\frac{10^n-1}{10-1}\right] = \frac{8}{9}\left[10^n-1\right]$$

So we get $$\displaystyle \left\{\frac{2}{3}\left[10^n-1\right]\right\}^2+\frac{8}{9}\left[10^n-1\right] = \frac{4}{9}(10^n-1)^2+\frac{8}{9}(10^n-1)$$

So we get $$\displaystyle = \frac{4}{9}(10^n-1)\cdot \left[10^n-1+2\right] = \frac{4}{9}(10^{2n}-1)$$

So we can write $$\displaystyle \frac{4}{9}(10^{2n}-1) = \underbrace{(444444444444\ldots)}_{2n~\text{times}}$$

Hint :

I will show it for 3-digit and you will know the general trend.

$(666)^2 + 888 = 36(111)^2 + 8(111) = 111(36(111) + 8) = 444(9(111) + 2) = 444(1001) = 444444$

Basically you will always get $44\ldots4$ ($n$ times) common times $100\ldots1$ ($n-1$ zeroes) which will help repeating the digits.

You can write a formal proof using this idea very easily. Let $e_n$ be the integer with $n$ $1s$. Substitute $e_n$ for $111$ in proof above and note that $9e_n+2=10^n+1$.