what is the units digit of the 99th number in the sequence: 1, 3, 9, 27, ...?

plz can some one tell  me the answer

Cphill plz can you tell me answer or Melody

Note  that the series can be  written as

3⁰, 3¹, 3², 3³.......  =

3⁴ⁿ, 3^{4n + 1},  3^{4n + 2},^, 3^{4n + 3}........  for n = 0, 1, 2, 3 ..........

And the last digit has the repeating pattern

1, 3, 9, 7 ...........

So....the 99th term  is   3⁹⁸  =   3^{4(24) + 2}  which will end in 9