Prove that every number of the sequence $49,4489,444889,\ldots$ is a perfect square

We can prove it as follows: \begin{align*} \underbrace{666\ldots6}_{n}\,7^2 &= \left[7 + 6(10^1 + 10^2 + \cdots + 10^n) \right]^2 \\ &= \left[7 + 6 \frac{10^{n+1} - 10}{9}\right]^2 \\ &= \left[ \frac{2 \cdot 10^{n+1} + 1}{3}\right]^2 \\ &= \frac{4 \cdot 10^{2n+2} + 4 \cdot 10^{n+1} + 1}{9} \\ &= 4 \cdot \frac{10^{2n+2} - 1}{9} + 4 \cdot \frac{10^{n+1} - 1}{9} + 1 \\ &= 4 \cdot \underbrace{111 \ldots 1}_{2n+2} + 4 \cdot \underbrace{111 \ldots 1}_{n+1} + 1 \\ &= \underbrace{444 \ldots 4}_{2n+2} + \underbrace{444 \ldots 4}_{n+1} + 1 \\ &= \underbrace{444 \ldots 4}_{n+1}\,\underbrace{888 \ldots 8}_{n}\,9. \end{align*}

The $n$th term is $T_{n}=9+8.10+8.10^2+...8.10^{n-1}+4.10^n+...+4.10^{2n-1}$:

$$ \begin{align} T_n &= 9+8.10(1+10+...10^{n-2})+4.10^n(1+...+10^{n-1}) \\ &= 9+8.10(\frac{10^{n-1}-1}{9})+4.10^n (\frac{10^{n}-1}{9}) \\ &=9+8.\frac{10^{n}-10}{9}+4.\frac{10^{2n}-10^n}{9} \\ &=\frac{81+8.10^n-80+4.10^{2n}-4.10^n}{9} \\ &=\frac{1}{9}.(1+4.10^n+4.10^{2n}) \\ &={(\frac{{1+2.10^n}}{3}})^2 \end{align} $$.

Now $10\equiv 1\pmod{3}\Rightarrow 10^n\equiv 1\pmod{3}\Rightarrow 2.10^n+1\equiv 0\pmod{3}$, thus $3$ divides the numerator and we are done.

Very similar to this question, which asks for a proof that $1111\ldots5556$ is a square. The proof is just as simple:

Multiply $4444\ldots8889$ by $9$ and you get $4000\ldots0004000\ldots0001$, which is $2000\ldots0001^2$.