Generating function of partition with restriction 
Let $c(m,n)$ denotes number of partitions $n$ into parts not greater than $m$, where order of elements does matter (so they are not classic partitions). Prove that:
  $$\sum_{n\ge 0}c(m,n)x^n = \frac{1-x}{1-2x+x^{m+1}}$$

I don't know how to approach. Generating function of such partitions is (if I'm not crazy): $(x+x^2+x^3+...+x^m)(x+x^2+x^3+...+x^m)(x+x^2+x^3+...+x^m)...$ but I don't see how it can help here.
 A: Note 1: This incorporates the correction in the comments below.
Note 2: I have incorporated  Marc van Leeuwen's suggestion about
simplifying the proof, though
I left in my original proof.
Working with your expression,
let
$f_m(x) =x+x^2+x^3+...+x^m
=x(1+x+x^2+x^3+...+x^{m-1})
=\frac{x(1-x^m)}{1-x}
$.
Then $f_m(x)$ is the GF (generating function)
of partitions into 1 term $\le$ m,
$f_m^2(x)$ is the GF 
of partitions into 2 terms $\le$ m,
and, in general,
$f_m^k(x)$ is the GF 
of partitions into $k$ terms $\le$ m.
Here is what I $should$ have done,
based on  Marc van Leeuwen's comment:
Let $F_m(x, u, v)$
be the GF 
of partitions into $u$ through $v$ terms (inclusive)
 $\le$ m.
The interesting case is, of course,
$u=0$ and $v = \infty$.
Then
$\begin{align}
F_m(x, u, v) 
&= \sum_{k=u}^{v} f_m^k(x)\\
&= f_m^u(x)\sum_{k=0}^{v-u} f_m^k(x)\\
&= f_m^u(x)\frac{1-f_m^{v-u+1}(x)}{1-f_m(x)}\\
\end{align}
$
If $u=0$,
$\begin{align}
F_m(x, 0, v) 
&= \frac{1-f_m^{v+1}(x)}{1-f_m(x)}\\
&= \frac{1}{1-f_m(x)}(1-f_m^{v}(x))\\
&= \frac{}{1-\frac{x(1-x^m)}{1-x}}(1-f_m^{v}(x))\\
&= \frac{1-x}{1-x-(x(1-x^m))}(1-f_m^{v}(x))\\
&= \frac{1-x}{1-2x+x^{m+1}}(1-f_m^{v}(x))\\
\end{align}
$
If $u=0$ and $v=\infty$,
$F_m(x) = F_m(x, 0, \infty) 
= \dfrac{1-x}{1-2x+x^{m+1}}
$
That is what I $should$ have done.
Instead, I did this:
So the total GF,
writing $f_m^0(x)=1$, is
$\begin{align}
F_m(x) &= 
\sum_{k=0}^{\infty} f_m^k(x)\\
&= \sum_{k=0}^{\infty} \big(\dfrac{x(1-x^m)}{1-x}\big)^k\\
&= \sum_{k=0}^{\infty} \big(\dfrac{x}{1-x}\big)^k\sum_{j=0}^k  \binom{k}{j}(-1)^j x^{jm}\\
&= \sum_{j=0}^{\infty}(-1)^jx^{jm}\sum_{k=j}^{\infty}\binom{k}{j} \big(\dfrac{ x}{1-x}\big)^k\\
\end{align}
$
Since
$\sum_{k=n}^{\infty}\binom{k}{n}x^{k} = \dfrac{x^n}{(1−x)^{n+1}}
$,
$\begin{align}
\sum_{k=j}^{\infty}\binom{k}{j} \big(\dfrac{ x}{1-x}\big)^k
&= \dfrac{ (x/(1-x))^j}{(1-(x/(1-x))^{j+1}}\\
&= \dfrac{ (x/(1-x))^j}{((1-2x)/(1-x))^{j+1}}\\
&= \dfrac{ x^j(1-x)}{(1-2x)^{j+1}}\\
&= \dfrac{ 1-x}{1-2x}\big(\dfrac{ x}{1-2x}\big)^{j}\\
\end{align}
$
Putting this in,
$\begin{align}
F_m(x) &= 
\sum_{k=1}^{\infty} f_m^k(x)\\
&= \sum_{j=0}^{\infty}(-1)^jx^{jm}\dfrac{ 1-x}{1-2x}\big(\dfrac{ x}{1-2x}\big)^{j}\\
&= \dfrac{ 1-x}{1-2x}\sum_{j=0}^{\infty}(-1)^jx^{jm}\big(\dfrac{ x}{1-2x}\big)^{j}\\
&= \dfrac{ 1-x}{1-2x}\sum_{j=0}^{\infty}(-1)^j\big(\dfrac{ x^{m+1}}{1-2x}\big)^{j}\\
&= \dfrac{ 1-x}{1-2x}\frac1{1+\frac{ x^{m+1}}{1-2x}}\\
&= \dfrac{ 1-x}{1-2x+ x^{m+1}}\\
\end{align}
$
This is the same as your expression,
once I put in the correction suggested.
