Computing the nth derivative using the binomial theorem. Let n be an element of the set of  natural numbers
Let $F(x)=x^2(1+x)^n$ and write $F^n$ for the nth derivative of the function $F$. 
Compute $F^n$ by applying the Binomial Theorem to $(1+x)^n$.
I don't understand the step where I need to find the derivative of these terms:
= $x^2 + {n \choose 1}x^3 + {n \choose 2} x^4 + .... + {n \choose n-1} x^{n+1} + x^{n+2} $
Even though I do understand how to show that for the first 3 terms the nth derivative is 0, but for the last two I have no clue??
Really appreciate any hints!!

Write it in decreasing powers of $x$ and use subscript $n$:
$$F_n(x)=x^2(1+x)^n=x^2(x+1)^n=\\
\color{red}{x^{n+2}}+\color{green}{nx^{n+1}}+\color{blue}{\frac{n(n-1)}{2}x^{n}}+\frac{n(n-1)(n-2)}{6}x^{n-1}+\cdots+nx^3+x^2$$
Note that for $f(x)=x^n$:
$$f^{(0)}(x)=f(x)=x^n\\
f^{(k)}(x)=\begin{cases}n(n-1)(n-2)\cdots(n-k+1)x^{n-k},k\le n\\ 0,k>n\end{cases}$$ 
Hence:
$$F_0(x)=x^2 \Rightarrow F_0^{(0)}(x)=F_0(x)=x^2;\\
F_n^{(n)}(x)=\color{red}{(n+2)(n+1)n\cdots3\cdot x^2}+\color{green}{n\cdot (n+1)n\cdots1\cdot x}+\\
\color{blue}{\frac{n(n-1)}{2}\cdot n(n-1)\cdots 1}=\\
\frac12(n+2)(n+1)\cdot n!\cdot x^2+n(n+1)\cdot n!\cdot x+\frac{n(n-1)}{2}\cdot n!=\\
n!\left(\frac12(n+2)(n+1)x^2+n(n+1)x+\frac{n(n-1)}{2}\right).$$
Verify:
$$F_1(x)=x^2(x+1)=x^3+x^2 \Rightarrow F_1^{(1)}(x)=1!(3x^2+2x)=3x^2+2x;\\
F_2(x)=x^2(x+1)^2=x^4+2x^3+x^2 \Rightarrow \\
F_2^{(2)}(x)=2!(6x^2+6x+1)=12x^2+12x+2.$$
