For every k dividing p-1, there are at most k kth roots of 1. For instance, set p = 37. There are 2 square roots, 1 and -1, which we can associate with 36 and 18, p-1 and half of p-1 respectively. Obviously 18 is not -1, not a square root of 1, I'm just associating roots with numbers for bookkeeping purposes. There are at most 4 fourth roots, two of them square roots, already accounted for, so associate the other two with 9 and 27, ¼ and ¾ of p-1. There are at most 3 cube roots, one of them is 1, so associate the other two with 12 and 24. There are 2 sixth roots that are not square roots or cube roots, associate them with 6 and 30. Do this for all the proper factors of 36. When we're all done, there is nothing associated with 1, or any of the other numbers coprime to 36. There is at least one v out there that is not a square root of 1, not a cube root of 1, not a fourth root of 1, etc, yet it is a 36th root of 1. This is a primitive root mod p.
Each primitive root generates all the nonzero values mod p. Let v be a primitive root and multiply it by itself again and again. If vj = vk, then v is a k-j root of 1, and we know that doesn't happen until the exponent reaches p-1. The powers of v step through all the nonzero elements mod p, though the order is impossible to predict in advance.
If s and p-1 are relatively prime then let t be the inverse of s mod p-1. When p is 37, and p-1 is 36, s might be 7, and t would be 31. when vs is raised to the t, the result is v again. The powers of vs cover all the powers of v, including v itself, hence vs is another primitive root. How many primitive roots are there? One for each exponent that is coprime to p-1. There are 12 primitive roots mod 37; the smallest is 2.
Field theory makes it easy to prove all of this, and the proof is more general than the one above, (it applies to more situations), but I'm not assuming any knowledge of fields here, and you're probably grateful.
Artin's conjecture says that every positive integer, not a square, is primitive for infinitely many primes.