According to the fundamental theorem of arithmetic, every integer greater than 1 can be represented uniquely as a product of prime numbers. Let's represent even exponents as \(e_i\) and odd exponents as \(o_i\). Example:
$$ n = {p_1}^{e_1} * {p_2}^{o_2} * {p_3}^{e_3} * {p_4}^{o_4} * {p_5}^{o_5} $$
Let's group all the factors with even exponents together, and the factors with odd exponents together:
$$ n = ({p_1}^{e_1} * {p_3}^{e_3}) * ({p_2}^{o_2} * {p_4}^{o_4} * {p_5}^{o_5}) $$
Since \(p^{3+b} = p^3 * p^b\):
$$ n = ({p_1}^{e_1} * {p_3}^{e_3}) * ({p_2}^3 * {p_4}^3 * {p_5}^3) * ({p_2}^{f_2} * {p_4}^{f_4} * {p_5}^{f_5}) $$
where \(f_i = o_i - 3\). We can group the even primes again, and also factor 3 out:
$$\begin{align} n &= ({p_1}^{e_1} * {p_3}^{e_3} * {p_2}^{f_2} * {p_4}^{f_4} * {p_5}^{f_5}) * ({p_2}^3 * {p_4}^3 * {p_5}^3) \\ n &= ({p_1}^{e_1} * {p_3}^{e_3} * {p_2}^{f_2} * {p_4}^{f_4} * {p_5}^{f_5}) * ({p_2} * {p_4} * {p_5})^3 \end{align}$$
All \(e_i\) and \(f_i\) are even, so we can factor 2 out. This shows that the powerful number can presented as a product of a square number and cube number:
$$ n = (\ldots )^2 * (\ldots )^3 $$