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