$9^1=9$
$9^2=81$ units digit $1$
$9^3=9\times81$
units digit comes from $9\times1=9$
$9^4=9\times9^3$
units digit comes from $9\times9$ so it will be $1$
And so on - units digits go $9,1,9,1,9,1...$
So the last digit of this sum will come from $9+1=10$ - so it will be $0.$