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.

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
$$
\begin{align}
\frac{1}{(1-x)^{k+1}}
&=(1-x)^{-(k+1)}\\
&=\sum_n\binom{-(k+1)}{n}(-x)^n\\
&=\sum_n(-1)^k\binom{n+k}{n}(-x)^n\\
&=\sum_n\binom{n+k}{n}x^n\tag{4}
\end{align}
$$
