By observing the following numbers

• $34^2 = 1156$
• $334^2=111556$
• $3334^2=111556$

we will prove an enhanced conclustion: $\underbrace{333\cdots 3}_{k}4^2=\underbrace{111\cdots 1}_{k+1}\underbrace{555\cdots 5}_{k}6$.

There are several ways to show this. One primitive method is just to do the calculation. We note that: $$3\times\underbrace{333\cdots3}_{k}4=3\times(\underbrace{333\cdots3}_{k+1} + 1) = \underbrace{999\cdots9}_{k+1} + 3 = 1\underbrace{000\cdots 0}_{k}2$$

(For example: $3\times 34 = 102$, $3\times 334=1002$, $\cdots$)

Then,$$\begin{align}&(\underbrace{333\cdots3}_{k+1})\times\underbrace{333\cdots3}_{k}4 \\=& (3 + 30 + \cdots + 3\times 10^{k})\times \underbrace{333\cdots3}_{k}4\\=&1\underbrace{000\cdots 0}_{k}2 + 1\underbrace{000\cdots 0}_{k}20 + 1\underbrace{000\cdots 0}_{k}200 +\cdots + 1\underbrace{000\cdots 0}_{k}2\underbrace{00\cdots 0}_{k}\\=&\underbrace{11\cdots 1}_{k+1}\underbrace{22\cdots 2}_{k+1}\end{align}$$

(For example: $33\times 34=1122$, $333\times 334=111222$, $\cdots$)

It follows that $$\begin{align}&\underbrace{333\cdots 3}_{k}4^2\\=&\underbrace{333\cdots 3}_{k}4\times \underbrace{333\cdots 3}_{k}4\\=& (\underbrace{333\cdots 3}_{k+1} + 1)\times \underbrace{333\cdots 3}_{k}4\\=&\underbrace{333\cdots 3}_{k+1}\times \underbrace{333\cdots 3}_{k}4 + \underbrace{333\cdots 3}_{k}4\\=&\underbrace{11\cdots 1}_{k+1}\underbrace{22\cdots 2}_{k+1}+\underbrace{333\cdots 3}_{k}4\\=&\underbrace{11\cdots 1}_{k+1}\underbrace{555\cdots 5}_{k}6\end{align}$$