Pascal's squared digit sum

We can see a pattern in each row of the Pascal's Triangle. If we take the $p$ th row of the Pascal's Triangle and the $r$ th element of that row, then the value of the element is equal to the binomial coefficient $\binom{p}{r-1}$ i.e, $pC(r-1)$ . So, in this case, the value of the $16$ th element in the $28$ th row is equal to the binomial coefficient $\binom{28}{16-1}=\binom{28}{15} = \frac{28!}{15!\times (28-15)!}=\frac{28!}{15!\times 13!}=\boxed{37442160}$

Now, we have that $n$ is the digit sum of this large value, so we have ----

$n=3+7+4+4+2+1+6+0=\boxed{27}$

So, finally we have $n^2=27^2=\boxed{729}$

$\color{#D61F06}{\Large\textbf{NOTE: 729 IS THE WRONG ANSWER.}}$

The author of this problem assumed that the row "1" is the first row, not the zeroth row, and so on. So when the author says "twenty-eighth row", he/she should mean "the row starting with 1,29,...", NOT "the row starting with 1,28,...".

At first I computed the correct answer, but it was > 1000, so I assumed the author made this mistake and got it correct.