Sequence of perfect squares

Prove that each term of the following sequence is a perfect square:

49, 4489, 444889, 44448889, 4444488889, 444444888889, …

In general, the nth term has n 4′s and n-1 8′s and one 9.

(Laars Helenius/FB/Math)

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Notation: <n> = 444…48…889 with n digits 4, etc (n>=1).
9·<n> = 10·<n> – <n> = 444…8…8890 – 444…4…8889
Here the first … in the terms mean 4:s, and the second … mean 8:s. The 8 in the first term and the 4 in the second one are in the same position (n digits to the right of them). Thus, 9·<n> = 4·10²ⁿ + 4·10ⁿ + 1 = (2·10ⁿ + 1)² where the last 1 comes from 90 – 89.

Now, 2·10ⁿ + 1 is obviously divisible by 3, i e 2·10ⁿ + 1 =3·m (m integer) so <n> = (3·m)²/9 = m², i e <n> is a square. – qed