What is the unit digit of $$(217)^{413} \times (819)^{547} \times (414)^{624} \times (342)^{812}$$?

Power series of 7 i.e units digit 7 power expansion has 7,9,3 and 1 and it is raised to power 413 i.e 413/4 remainder 1 and so last digit is 7
Power series of 9 i.e units digit 9 power expansion has 9 and 1 and it is raised to power 547 i.e 547/2 remainder 1 and so last digit is 9
Power series of 4 i.e units digit 4 power expansion has 4 and 6 and it is raised to power 624 i.e 624/2 remainder 0 and so last digit is 6
Power series of 2 i.e units digit 2 power expansion has 2,4,8 and 6 and it is raised to power 812 i.e 812/4 remainder 0 and so last digit is 6
All the last digits product=7*9*6*6
=8