The part I don't understand is if the p equal instalments are paid at times $1/p, 2/p, 3/p, \dots , 1$ then how does the final line get the power of $(p-1)/p, (p-2)/p, \dots , 1$? i.e. should the last line not read: $$\dfrac{i^{(p)}}{p}(1+i)^{1/p} + \dfrac{i^{(p)}}{p}(1+i)^{2/p} + \dots + \dfrac{i^{(p)}}{p} = i$$?
When an installment is paid at time $k/p$, the amount of time that remains for that installment to accrue interest is $$1 - \frac{k}{p} = \frac{p-k}{p}.$$ So if the payments are made at $1/p, 2/p, 3/p, \ldots, p/p$, then the amount of time that they accrue are, repsectively, $$\frac{p-1}{p}, \frac{p-2}{p}, \frac{p-3}{p}, \ldots, \frac{p - p}{p},$$ where the last one is zero.
At this point, you should observe that in your expression, the value of your exponents is in increasing order up to the last term, which has an exponent of zero: $(1+i)^0 = 1$. This is why you are getting confused, because you are basically writing out the sequence $1, 2, 3, \ldots, 0$, and you've not properly specified how the penultimate term should be, namely, where it stops. Should it be $p$, or $p-1$? Some thought indicates it should be the latter, since there are $p$ installments in total. So if you need to write it your way, you would do better to write $$\frac{i^{(p)}}{p} \left( (1+i)^0 + (1+i)^{1/p} + (1+i)^{2/p} + \cdots + (1+i)^{(p-1)/p}\right) = i.$$ And now, you can see that this is actually equivalent to the textbook's expression, simply written in the reverse order, since if you write out more terms on both ends of the sum, it is really $$\frac{i^{(p)}}{p} \left( (1+i)^0 + (1+i)^{1/p} + (1+i)^{2/p} + \cdots + (1+i)^{(p-3)/p} + (1+i)^{(p-2)/p} + (1+i)^{(p-1)/p}\right) = i.$$