Intuitive explanation for a polynomial expansion? Is there an ituitive explanation for the formula: 
$$
\frac{1}{\left(1-x\right)^{k+1}}=\sum_{n=0}^{\infty}\left(\begin{array}{c}
n+k\\
n
\end{array}\right)x^{n}
$$
 ?
Taylor expansion around x=0
 :
$$
\frac{1}{1-x}=1+x+x^{2}+x^{3}+...
 $$
differentiate this k
  times will prove this formula. but is there an easy explanation for this? Any thing similar to the binomial law to show that the coefficient of $x^{n}$
  is $\left(\begin{array}{c}
n+k\\
n
\end{array}\right)$
 .
Thanks in advance.
 A: If by the binomial law you mean
$$
(1+x)^n=\sum_k\binom{n}{k}x^k\tag{1}
$$
then yes. Note that
$$
\binom{n}{k}=\frac{n(n-1)(n-2)\dots(n-k+1)}{k!}\tag{2}
$$
Consider what $(2)$ looks like for a negative exponent, $-n$:
$$
\begin{align}
\binom{-n}{k}
&=\frac{-n(-n-1)(-n-2)\dots(-n-k+1)}{k!}\\
&=(-1)^k\frac{(n+k-1)(n+k-2)(n+k-3)\dots n}{k!}\\
&=(-1)^k\binom{n+k-1}{k}\tag{3}
\end{align}
$$
Plug $(3)$ into $(1)$ and we get
$$
\begin{align}
\frac{1}{(1-x)^{k+1}}
&=(1-x)^{-(k+1)}\\
&=\sum_n\binom{-(k+1)}{n}(-x)^n\\
&=\sum_n(-1)^k\binom{n+k}{n}(-x)^n\\
&=\sum_n\binom{n+k}{n}x^n\tag{4}
\end{align}
$$
A: Here’s one way of looking at it.
Suppose that you have $$f(x)=\sum_{n\ge 0}a_nx^n\;,$$ and you multiply both sides by $\frac1{1-x}$:
$$\begin{align*}
\left(\frac1{1-x}\right)f(x)&=\left(\sum_{n\ge 0}x^n\right)\left(\sum_{n\ge 0}a_nx^n\right)\\
&=(1+x+x^2+\dots)(a_0+a_1x+a_2x^2+\dots)\\
&=a_0 +(a_0+a_1)x+(a_0+a_1+a_2)x^2+\dots\\
&=\sum_{n\ge }\left(\sum_{k=0}^na_k\right)x^n\;.
\end{align*}$$
In other words, the coefficient of $x^n$ in the product is just $a_0+a_1+\dots+a_n$.
Now think about the construction of Pascal’s triangle:
$$\begin{array}{c}
1\\
1&1\\
1&2&1\\
1&3&3&1\\
1&4&6&4&1\\
1&5&10&10&5&1\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots
\end{array}$$
The binomial coefficient $\binom{n}k$ is the entry in row $n$, column $k$ (numbered from $0$). Moreover, since $$\binom{n}k=\binom{n-1}k+\binom{n-1}{k-1}\;,$$ each entry is the sum of the numbers above it in the column immediately to the left: this is the identity $$\sum_{i=0}^n\binom{i}k=\binom{n+1}{k+1}\;.\tag{1}$$
In the first ($k=0$) column of Pascal’s triangle you have the coefficients in the power series expansion of $\frac1{1-x}$. We saw above that the coefficients in the power series expansion of $\frac1{(1-x)^2}$ are just the cumulative sums of these coefficients, $1,2,3,\dots$, but these are just the entries in the second ($k=1$) column of Pascal’s triangle. Similarly, the coefficients in the power series expansion of $\frac1{(1-x)^3}$ are the cumulative sums of $1,2,3\dots$, or $1,3,6,\dots$, the numbers in the third ($k=2$) column of Pascal’s triangle. In general, the coefficients in the power series expansion of $\frac1{(1-x)^{k+1}}$ must be the binomial coefficients in the $k$ column of Pascal’s triangle, those of the form $\binom{n}k$. All that remains is to get the row indexing right: we want the $1$ that is the first non-zero entry in column $k$ to be the constant term. It’s in row $k$, so the coefficient of $x^n$ must in general be the binomial coefficient in row $n+k$, and we get 
$$\frac1{(1-x)^{k+1}}=\sum_{n\ge 0}\binom{n+k}kx^n=\sum_{n\ge 0}\binom{n+k}nx^n\;.$$
A: Consider the number of solutions (say $\displaystyle a_n$) to the equation:
$$x_1 + x_2 + \dots + x_{k+1} = n$$
Where $x_i$ are non-negative integers, and $n$ is a non-negative integer.
The Stars and Bars approach: choosing where to place $\displaystyle k$ bars, out of a possible $\displaystyle n+k$ spots, gives us that the number of solutions is exactly $a_n = \displaystyle \binom{n+k}{n}$
But, if you look at this using the Generating Functions approach, we see that
$$(1+x + x^2 + \dots)^{k+1} = \sum_{n=0}^{\infty} a_n x^n$$
i.e.
$$\frac{1}{(1-x)^{k+1}} = \sum_{n=0}^{\infty} a_n x^n = \sum_{n=0}^{\infty} \binom{n+k}{n} x^n $$
