To Prove:

For all n in the Natural Numbers, v(n) = k if and only if 10^(k-1) <= n < 10^k

v(n) is the number of digits of n with respect to base 10....so for example, v(d) = 1 for all digits d, and v(10) = 2.

I can't seem to figure out how to work this proof. Any help would be appreciated!