I feel as though this is fairly straightforward, but I can't figure it out.
If I have $n$ numbers, clearly these can be arranged in $n!$ ways. But if each of the $n$ numbers can have a value $v, \text{where } 0\le v<k$ then each has $k$ possibilities. If $n_1 := k'$, this says nothing of whether $n_2$ may also be $k'$.
In the case $k = 10$, this is the decimal number system; so the number of combinations is $10^n -1$.
Can this be generalised to $k^n - 1$? I'm suspicious at the lack of any factorial.