Finding the closed form of a generating function Given that $k$ is a positive integer and $f(x)$ is the generating function of the sequence $(b_0,b_1,b_2,...)$ where $b_n = {n \choose k}\;\, \forall \;n$, show that: $$f(x)=\frac{x^k}{(1-x)^{k+1}}$$
I tried writing a few terms of $f(x)$:$$f(x)=x^k+{k+1 \choose k}x^{k+1}+{k+2 \choose k}x^{k+2}+...$$$$\;\;\;\;\;\;\;\;\;=x^k\left(1+{k+1 \choose k}x^1+{k+2 \choose k}x^2+...\right)$$$$=x^k\cdot h(x)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; $$
Where $h(x)$ is defined as the expression in parenthesis, then I tried some manipulation, for example I computed $h(x)-xh(x)$, since: ${k+n+1 \choose k}-{k+n \choose k}={k+n \choose k-1}$ we have: 
$h(x)-xh(x)=1+{k \choose k-1}x+{k+1 \choose k-1}x^2+...$
But I'm stuck, not sure how I should procede

HINT:
Inductive hypothesis
\begin{eqnarray*}
f_k(x)= \sum_{n=k}^{\infty} \binom{n}{k} x^n = \frac{x^k}{(1-x)^{k+1}} 
\end{eqnarray*}
and use
\begin{eqnarray*}
 \binom{n}{k} + \binom{n}{k+1}= \binom{n+1}{k+1}.
\end{eqnarray*}
More detail available on request.
Further Hint :
\begin{eqnarray*}
f_{k+1}(x) &=& \sum_{n=k+1}^{\infty} \binom{n}{k+1} x^n = \sum_{n=k}^{\infty} \binom{n+1}{k+1} x^{n+1} \\
&=&  \sum_{n=k}^{\infty} \binom{n}{k} x^{n+1} + \sum_{n=k}^{\infty} \binom{n}{k+1} x^{n+1} \\ \cdots
\end{eqnarray*}
