KöMaL 2005 C817

Having calculated that $6^2$+8=44, Clare noticed that $66^2$+88=4444 was also true. Is it true for every $n$ that $(\underbrace{6\dots6}_{n\text{ digits}}{)}^2+ \underbrace{8\dots8}_{n\text{ digits}}=\underbrace{4\dots4}_{2n\text {digits}}$?