contest math – find the smallest positive multiple \$n\$ of 1994 such that the produc of the divisors of \$n\$ is \$n^{1994}\$

Find the smallest positive multiple $$n$$ of 1994 such that the produc of the divisors of $$n$$ is $$n^{1994}$$

Attempt

Let $$nin mathbb{N}$$ and denote by $$d(n)$$ the number of divisors of $$n$$ and $$p(n)$$ the product of all the divisors of $$n$$ including $$n$$.

Consider the factorization of $$n$$ given by $$n=p_1^{alpha_1} p_2^{alpha_2}cdots p_n^{alpha_n}$$ for $$alpha_iin mathbb{N}$$. Is known the fact that $$p(n)=n^{d(n)/2}=n^{1994}$$ it is
$$d(n)=997 cdot 2^2$$ in addition $$n=997 cdot 2 cdot k,$$ for some $$kin mathbb{N}$$.

since $$n$$ is the smallest multiple of $$1994$$ with $$997 cdot 2^2$$ divisors then
$$n$$ should be $$2^{998} cdot 997$$

Any comment or corretion was helpful