Proof of $C\left(n, k\right) = \frac{C\left(n, k-1\right) \cdot C(n - (k - 1), 1)}k$ I was wondering if anyone could explain to me why this holds true?

$$C\left(n, k\right) = \frac{C\left(n, k-1\right) \cdot C(n - (k - 1), 1)}k$$

or

$$C\left(3, 2\right) = \frac{C\left(3, 1\right) \cdot C(2, 1)}2$$

I was playing around with combinatorics, and realized that this holds true for all values that I put into it.
Here is a combinatorical explanation:

You can choose $k$ from $n$ items by


choosing $k-1$ from $n$: $\color{blue}{\binom{n}{k-1}}$ and then
choosing $1$ from the remaining $n-(k-1)$: $\color{blue}{\binom{n-(k-1)}{1}}$


But using this selection method the same selection can result in $\color{blue}{k}$ ways, since the $1$ you choose "last" can be one of the $k$. So,

$$\color{blue}{\binom{n}{k-1}\cdot\binom{n-(k-1)}{1}} = \color{blue}{k}\cdot \binom{n}{k}$$
LHS is the number of ways to choose $k$ objects from $n$ objects. 
RHS is nothing but, now from $n$ objects first choose $k - 1$ and then from remaining $n-(k-1)$ objects choose 1 obeject. 
but in this counting, we are repeating the same objects to objects K, so divided by Y
enter image description here

$\frac { n ! } { k ! ( n - k ) ! } \simeq \frac { 1 } { k } \frac { n ! } { ( n - 1 ) ( n - k + 1 ) ! } \frac { ( n - k + 1 ) ! } { 1 ! ( n - k ) ! }$