An interesting property of binomial coefficients that I couldn't prove So when I was trying to prove the argument in this link I've come up with something.
When you  extract the left term from the right term, you get the term under them.
What is interesting is that as you can see it follows a pattern similar to binomial coefficients
such $1,2,1 - 1,3,3,1 - 1,4,6,4,1$ etc. and when $k=0$ in the first layer, all terms are equal to $0$
and when $k=1$ in the second layer all terms are equal to $0$, and when $k = 2$ in the third layer
all terms are equal to $0$ and so on.
$1^k-0^k \qquad\qquad 2^k-1^k \qquad\qquad 3^k-2^k \qquad\qquad 4^k-3^k$
$\qquad 2^k-2.1^k+0^k \qquad 3^k-2.2^k+1^k \qquad 4^k-2.3^k+2^k$
$\quad\quad\quad 3^k-3.2^k+3.1^k-0^k \quad\quad 4^k-3.3^k+3.2^k-1^k$
$\qquad \qquad \qquad 4^k-4.3^k+6.2^k-4.1^k+0^k$
So, I really couldn't figure out why. If I can prove it I will be able to prove the argument in the link I posted.
 A: This is what happens when you apply finite differences to any sequence. Here is some useful notation. If $a_n$ is a sequence, its forward difference is the sequence
$$\Delta a_n = a_{n+1} - a_n.$$
(The notation should not be read as "$\Delta$ of $a_n$," but as "the $n^{th}$ term of the sequence $\Delta a$.") For example, if $a_n = n^2$, then
$$\Delta a_n = (n+1)^2 - n^2 = 2n + 1.$$
The reason this notation is so useful is that it can be iterated: we can inductively define
$$\Delta^k a_n = \Delta^{k-1} a_{n+1} - \Delta^{k-1} a_n$$
and these are the differences of the differences. For example,
$$\Delta^2 a_n = (a_{n+2} - a_{n+1}) - (a_{n+1} - a_n) = a_{n+1} - 2 a_{n+1} + a_n$$
and
$$\Delta^3 a_n = a_{n+3} - 3 a_{n+2} + 3 a_{n+1} - a_n.$$
In general, it turns out that
$$\Delta^k a_n = \sum_{i=0}^k (-1)^i {k \choose i} a_{n+k-i}$$
and this is the binomial coefficient pattern you observe. You can prove this by induction, but to my mind, the cleanest way to prove it - the way that lets you "see it at a glance" - is to use the concept of operators. 
"Forward difference" is an operator: it eats up a sequence and spits out a sequence, kind of like differentiation, which eats up a function and spits out another function. Operators form a ring, an infinite-dimensional generalization of rings of matrices, and in particular you can add and multiply them (addition is pointwise, multiplication is composition). 
The $k^{th}$ forward difference operator is then literally the product $\Delta^k$ of $k$ copies of the forward difference operator in this ring. The significance of this observation is that $\Delta$ can be written as a difference of two operators, the identity operator $I$, which does nothing:
$$I a_n = a_n$$
and the forward shift operator, which shifts a sequence forward:
$$S a_n = a_{n+1}.$$
The precise relationship is that $\Delta = S - I$, and using the binomial theorem we can now write
$$\Delta^k = (S - I)^k = \sum_{i=0}^k {k \choose i} (-1)^i S^{k-i}$$
which is exactly the desired result. 
