Suppose $i^{(p)}$ denotes the nominal interest rate, payable as $\frac{i^{(p)}}{p}$, at a rate of $p$ times per period, and assume that $i$ is the effective interest rate (payable once per period) that is equivalent.
Then, by considering the accumulated value of the interest payments made under the nominal rate, we deduce that
$$ \frac{i^{(p)}}{p} + \frac{i^{(p)}}{p}(1+i) + \frac{i^{(p)}}{p}(1+i)^2 + \dots + \frac{i^{(p)}}{p}(1+i)^\frac{p-1}{p} = i \hspace{10mm} (*) $$
Then, by using the rule $$ \sum_{k=0}^{n-1}ar^k = a \cdot \frac{(1-r^n)}{1-r} $$ and taking $$ a = \frac{i^{(p)}}{p}, \hspace{10mm} r = (1+i)^\frac{1}{p}, \hspace{10mm} n = p $$ we can express $(*)$ as $$ \frac{i^{(p)}}{p} \cdot \frac{1 - \left( (1+i)^\frac{1}{p} \right)^p}{1 - (1+i)^\frac{1}{p}} = \frac{i^{(p)}}{p} \cdot \frac{i}{1 - (1+i)^\frac{1}{p}} = i $$ Thus, by dividing through by $i$ we have $$ \frac{i^{(p)}}{p[1 - (1+i)^\frac{1}{p}]} = 1 $$ Giving the result $$ i^{(p)} = p[1 - (1+i)^\frac{1}{p}] $$
This result is incorrect, since the correct result should be $$ i^{(p)} = p[(1+i)^\frac{1}{p}-1] $$
Where in my attempted derivation did I go wrong?
You've forgotten a minus sign:
$$ \frac{i^{(p)}}{p} \cdot \frac{1 - \left( (1+i)^\frac{1}{p} \right)^p}{1 - (1+i)^\frac{1}{p}} = \frac{i^{(p)}}{p} \cdot \frac{-i}{1 - (1+i)^\frac{1}{p}} = i $$