Just checking if an answer is correct permutations

Question

Given the digits 0, 1, 2, 3, 4, 5, 6 and 7, how many valid seven digitnumbers can be made

if there is no repetition of digits

and a seven digit number may not start with zero?

This is a permutation because order does not matter???

The answer I have is 7*7*6*5....

or 7*7P6 = 35280

Answer 1

This is a permutation because order matters. The first digit cannot be 0, so you have 7 choices. Then you have 7 digits remaining since you cannot repeat digits. You have to choose 6 without repeating and the order matters, so this is a permutation.

$7 * 7P6 = 7 * \frac{7!}{(7-6)!} = 35280$