Hi Melody. The n in (2n+1) is just an integer.
For a square root there are two solutions, so n takes on two values, say 0 and 1, and 2n+1 becomes 1 and 3.
These give two values of -1 when plugged in to ei(2n+1)pi. The two square roots of -1 are then found by taking the square root of these (ei(2n+1)pi)1/2 = ei(2n+1)pi/2. When n = 0, ei(2n+1)pi/2= eipi/2 = i, and when n = 1, ei(2n+1)pi/2= ei3pi/2 = -i. Hence we have two values (i and -i) for the square root of -1, one for each value of n.
Similarly, we need three values of n (0, 1 and 2) for finding the three cube roots of -1, and four values for the four fourth roots etc.