Why there are $n^{n}$ possible arrangments? I found this problem:
Suppose $n$ balls are distributed into $n$ boxes so that all of
the $n^{n}$ possible arrangements are equally likely. Compute the probability
that only box 1 is empty.
Why there are $n^{n}$  possible arrangments?
If for the first box can be $0,1,2...,n$ balls($n+1$ possibilities).
For the second box $0,1,2,...,n-1$ balls ($n$ possibilities)
And so on..
So it must be $(n+1)!$ possible arrangements. Right?
 A: To explain the $n^n$ thing, I suggest this way. We'll build a PROCESS, then apply the multiply rule. Notice that the result you provide only true when all the balls are distinguishable and all the boxes are also distinguishable.
Since each of the balls must be arranged in a box, we'll build a process, in which each ball will be put in the boxes in an order. N
There are $n$ ways to put the first ball into the boxes (you can put it in the first box, the second box,... or the n-th box)
Similarly, there are $n$ ways to put the second ball, ... and the n-th ball into the boxes.
Since the way you put the i-th ball is not related to the way you put the j-th ball, thus we can apply the multiply rule, and get the desired result
P.s: I might have made some lexical mistakes, as my English is not good.
A: "$n$ balls distributed into $n$ boxes" has a bunch of different interpretations.
a) uniformly randomly throw undistinguishable  balls into $n$ distinguishable boxes: for each ball you have $n$ choices, for $n$ balls they total to $n^n$.
a')uniformly randomly throw distinguishable  balls into $n$ distinguishable boxes: same as before.  
That's because by throwing we are normally assigning a equiprobable space to the events ($k$th ball picked, $j$th box hit) and not to the resulting configuration (ball,box): actually some of these latter configurations will be repeated. So it is like we pick a ball and mark it according to the lauching sequence (whether or not it was already marked by other means).  
b) uniformly random filling (I don't know the standard denomination) of un/distinguishable boxes with un/distinguishable balls. Now the equiprobable events are considered to be the each "different configuration" of the (ball,boxes), which depend on the four mixing  possibilities of un/distinguishable.
Note that the case b), with distinguishable (marked) balls and distinguishable boxes does not correspond to a), since we pass from b-dd) to a) by ranking the balls in each box according to their marking: the ball launched first will always be lower in the box (figured as a can of piled balls).
