There is a unique such representation:
$$15! = 28 \times 27 \times 26 \times 25 \times 22 \times 21 \times 20 \times 18 \times 16.$$
All solutions start by factorising $15!$ into primes:
$$15! = 2^{11} \times 3^6 \times 5^3 \times 7^2 \times 11 \times 13.$$
This can be obtained manually or by Legendre's formula for the highest power of $p$ that divides $n!$ viz.
$$\left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \dots$$
Systematic Solution
As there is only one multiple of $11$ and one multiple of $13$ between $16$ and $30$ inclusive, two of the factors in the product must be $22$ and $26$.
There are two multiples of $7$, and no multiples of $7^2$ between $16$ and $30$, so both $21$ and $28$ must feature in the product.
There are three multiples of $5$ between $16$ and $30$, one of which is $5^2$. So the multiples of $5$ in the product must be either $20$ and $25$, or $25$ and $30$.
Taking out the factors already considered:
$$\frac{15!}{22 \times 26 \times 21 \times 28 \times 25} = 2^7 \times 3^5 \times 5 = 155520.$$
To finish the problem we seek four further factors between $16$ and $30$ (inclusive) excluding those already used.
Of the remaining numbers between $16$ and $30$, only $\{18, 24, 27, 30\}$ are multiples of $3$. If we do not choose $27$, then we can reach at most $3^4$ in the product, which is short of the $3^5$ we need. Therefore, $27$ must be in the list of factors. Dividing through by $27$, we are left to find three factors of:
$$\frac{15!}{22 \times 26 \times 21 \times 28 \times 25 \times 27} = 2^7 \times 3^2 \times 5 = 5760.$$
The feasible factors must only have prime factors of $2$, $3$ and/or $5$. Excluding factors already used, we have the following list:

Start with $15!$ and starting from $30$, divide repetitively by the largest remaining number that is a factor.
Taking the largest possible number at each stage
$$15! = 30 \times 28 \times 27 \times 26 \times 25 \times 24 \times 22 \times 21 \times 8$$
This has nearly worked, but the final factor of $8$ is problematic.
So we tweak a bit, first replacing $8$ and $30$ by $20$ and $12$
$$15! = 28 \times 27 \times 26 \times 25 \times 24 \times 22 \times 21 \times 20 \times 12$$
This is still not a solution, but our smallest factor has increased a bit. As a final step, replace $24$ and $12$ by $18$ and $16$ which gives the solution stated.