Is there a natural number whose product of all true divisors equals itself?

Is there a natural number whose product of all true divisors equals itself?

Of course, 6 is a good example. Its true divisor is 1,2,3
The natural number obtained by multiplying any two different prime numbers satisfies that the product of the true divisor is equal to itself, such as 10 = 2 * 5, 15 = 3 * 5, etc
The theory is: there are two different prime numbers u, V, whose product is m = u * v. according to the definition of true divisor, the true divisor of M is 1, u, V, so the product of its true divisor is 1 * u * V = u * V = M