How to simplify $C(k, 2)$ combinatorics 
$$\frac{k!}{2!(k-2) !} = \frac{1}{2}k(k-1)$$

I am having trouble figuring out how they arrived at this "simplification". This is from a book I am reading and they are using binomial distribution to prove that the derivative of $f(x) = x^k$ is $f'(x) = kx^{k-1}$.
While I understand the whole concept that the book is trying to get across, I am scared I might have to 'puke out" the exact formula (which contains the expression on the right, $\frac{1}{2k(k-1)}$ and I'd prefer understanding how they simplified it because understanding > memorizing
I am sorry if it's super obvious to most of you, I am quite bad with fractions and just recently got into mathematics... 
 A: The defining property of factorials is that for any $n$, $n!=n\cdot (n-1)!$.  We now apply this property twice, once for $n=k$ and again for $n=k-1$, to get:
$$k!=k\cdot (k-1)! = k\cdot (k-1)\cdot (k-2)!$$
A: Your equation is a specific case of the more general:
$${k\choose n} = \frac{k!}{n!(k-n)!} = \frac{k\cdot(k-1)\cdot ...\cdot (k-n+1)}{n!}$$
because:
$$\frac{k!}{n!(k-n)!} = \frac{k\cdot(k-1)\cdot ...\cdot (k-n+1) \cdot (k-n)!}{(k-n)!n!}=\frac{k\cdot(k-1)\cdot ...\cdot (k-n+1)}{n!}$$
A: Do it combinatorially: $C(k,2)$ counts the number of ways to choose, in no particular order, $2$ objects from a box of $k$ (distinct) objects. For the first of your pair of objects, there are $k$ options. For the second you have $k-1$ choices. The order doesn't matter, so we don't regard choosing the first object and then the second as any different than choosing the second and then the first. Then there are $k\cdot (k-1)/2$ such choices in total. 
A: $$C_{n}^{k}=\frac{n!}{k!(n-k)!}$$
$$C_{n}^{2}=\frac{n!}{2!(n-2)!}=\frac{1*2*3*....*(n-3)*(n-2)*(n-1)*n}{1*2*1*2*3*...*(n-4)*(n-3)*(n-2)}$$
When you remove the unnesessary things you get the identity you wanted:
$$C_{n}^{2}=\frac{n*(n-1)}{2}$$
Example $$C_{4}^{2}=\frac{3*4}{2}$$
