probability – What’s more likely: $7$-digit number with no $1$’s or at least one $1$ among its digits?

A $7$-digit number is chosen at random. Which is more likely: the number has no $1$‘s among its digits or the number has at least one $1$ among its digits?

Here’s how I did it: The question is asking whether $8(9)^6$ (the number of those with no $1$‘s among its digits) or $9(10)^6 – 8(9)^6$ (the number of those with at least one $1$ among its digits). Some tedious multiplying shows that $8(9)^6 = 4241528 < 4500000$, which demonstrates that $9(10^6) – 8(9)^6$ i.e. the number having at least one $1$‘s among its digits is more likely.

However, I am wondering if there is a slicker way to get the answer without having to do any tedious multipication.