Order of a number 
Does there always exist some integer $a$ such that $a^{p-1} \equiv 1 \pmod{p}$ where $p$ is a prime and $p-1 = \text{ord}_{p}(a)$?

I was wondering about this and it is basically saying, "can the maximal possible order always be achieved?" We know that it is always possible to achieve $p-1$, but is it possible to achieve it so that it is the minimal possible power such that $a^{p-1} \equiv 1 \pmod{p}$?
Edit: This leads me onto another question. Since such an $a$ must exist, is it true that all orders of integers $m$ modulo $p$ must divide $p-1$?

To answer your second question, a general fact from group theory, known as Lagrange's theorem, implies that the order of every element divides the size of the group, which in this case is $p-1$.
