Ones

The pattern is $(\underbrace{ 111...111}_n)^2=\underbrace{111...111}_{n-1}\underbrace{000...000}_n1$ . It always holds true and it is able to prove it.

Any $(\underbrace{111...111}_n)$ always equals to $(1\underbrace{000...000}_n-1)$

Therefore, $(\underbrace{111...111}_n)^2\\=(1\underbrace{000...000}_n-1)^2\\=(1\underbrace{000...000}_n)^2-1\underbrace{000...000}_{n+1}+1\\=1\underbrace{000...000}_{2n}-1\underbrace{000...000}_n1\\= \underbrace{111...111}_{n-1}\underbrace{000...000}_n1$ .

By the same method, we can get a conclusion:

For any numbers $(\underbrace{aaa...aaa}_n) _{a+1}$ , $(\underbrace{aaa...aaa}_n) _{a+1}^2$ always equals to $\underbrace{aaa...aaa}_{n-1}\underbrace{000...000}_n1_{a+1}$

Back to the question, $11111\times 11111=\boxed{1111000001}$

We a see a pattern from the given so we can conclude that $\implies 11111\times 11111 = 1111000001$