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This is mighty tricky, and the answer sheet (attached and also available on the OCW site) has a typo - k inside the binomial (n k) should be p.
If you apply Leibniz formula, you'll find that the first term is:
\[\left(\begin{matrix}n \\ 0\end{matrix}\right) (x^p)^{(p +q)}(1+x)^q \]
The thing to notice is that if you derive x^p more than p times, the derivative is going to be zero. Since you're deriving it p+q times, the derivative is going to be zero.
A similar argument can be applied to all terms except one, the term that is:
\[\left(\begin{matrix}n \\ p\end{matrix}\right) (x^p)^{(p)}[(1+x)^q]^{(q)} \]
Therefore, all the terms except the one above are zero.
Taking the derivative of that term is not really pretty, but from the end of lecture 4 (or from trying it yourself) you'll find that the derivative of
\[(x^p)^{(p)} = p! \]
Similarly, using the chain rule (try it with q = 2 or q =3 so you can see it)
\[[(1+x)^q]^{(q)} = q! \]
So the nonzero term is
\[\left(\begin{matrix}n \\ p\end{matrix}\right) p!q! \]
There's one more step: you have to note that n = p+q, so you can solve the binomial.
You'll get (for the binomial)
\[ \left(\begin{matrix}n \\ p\end{matrix}\right) = \frac{(p+q)!}{p!(p+q-p)!} = \frac{(p+q)!}{p!q!} = \frac{n!}{p!q!}\]
Substituting this in the nonzero term will get you:
\[ \frac{n!}{p!q!}p!q! = n!\]
This was kinda hard, but I hope it helped.

Oops, I forgot to substitute the n's in the (n 1) etc. terms, should instead be ((p+q) 1), etc.

Ok, I calculated it all out, and it works! Thanks!!!