Finalising proof from Humphreys´ "Introduction to Lie Algebras and Representation Theyory" $L=\mathfrak{sl}(2, \mathbb{F})$ with standard Chevalley basis $(x, \ y, \ h)$ and $a, \ c\in \mathbb{Z}^{+}$. Humphrey gives a Lemma in chapter 26.2 saying:
$\frac{x^{c}}{c!}\frac{y^{a}}{a!}=\sum\nolimits_{k=0}^{min\{a,c\}} \frac{y^{a-k}}{(a-k)!} \begin{pmatrix} h-a-c+2k \\ k \end{pmatrix} \frac{x^{c-k}}{(c-k)!}$
But only proves half of it.
He showes via induction that it works for $a$, when $c=1$, by proveing $\frac{xy^{a}}{a!}=\frac{y^{a}x}{a!}+\frac{y^{a-1}}{(a-1)!}(h-a+1)$.
The rest he only describes:
One has to use induction on $c$ utilising that $xf(h)=f(h-2)x$ for $f(T)$ polynomal and $\begin{pmatrix} n+1 \\ k \end{pmatrix} - \begin{pmatrix} n \\ k \end{pmatrix} = \begin{pmatrix} n \\ k-1 \end{pmatrix}$ to finish the proof. I am stuck with the induction:
The intitial step was given, with proving the $a$ part.
I tried the following for the induction step: $c \to c+1$
$\frac{x^{c+1}}{(c+1)!}\frac{y^{a}}{a!}  \\
= \frac{x}{c+1}(\frac{x^{c}y^{a}}{c!a!}) \\ 
= \frac{x}{c+1}(\sum\nolimits_{k=0}^{min\{a,c\}} \frac{y^{(a-k)}}{(a-k)!}\begin{pmatrix} h-a-c+2k \\ k \end{pmatrix} \frac{x^{(c-k)}}{(c-k)!})  \\
= \frac{1}{c+1} (\sum\nolimits_{k=0}^{min\{a,c\}} \frac{y^{(a-k)}}{(a-k)!} \begin{pmatrix} h - a - c + 2(k-1) \\ k \end{pmatrix} \frac{x^{(c-(k-1))}}{(c-k)!}) \\
= \dots$
But I have no clue how to continue.
By now I have calculeted some more:
First of all instead of the last step it is:
$\frac{1}{c+1}(\sum\nolimits_{k=0}^{min(c,a)} \frac{y^{a-k}}{(a-k)!} \begin{pmatrix} h-a-c+2(k-1) \\ k \end{pmatrix} \frac{x^{c-(k-1)}}{(c-k)!})+\frac{1}{c+1}(\sum\nolimits_{k=0}^{min(c,a)}\frac{y^‌​{a-(k‌+1)​}}{(a-(k+1))!}\begin{pmatrix}h-a+k+1\end{pmatrix} \begin{pmatrix} h-a-c+2k \\ k \end{pmatrix} \frac{x^{c-k}}{(c-k)!})$
Also I have tried doing things back to front:
$\sum_{k=0}^{min(a,c+1)}\frac{y^{a-k}}{(a-k)!}\begin{pmatrix} h-a-c+2k-1 \\ k \end{pmatrix}\frac{x^{c-(k-1)}}{(c-(k-1))!}  \\
=\sum_{k=0}^{min(a,c+1)}\frac{y^{a-k}}{(a-k)!}\left( \begin{pmatrix} h-a-c+2(k-1) \\ (k-1)\end{pmatrix} - \begin{pmatrix} h-a-c+2(k-1) \\ k \end{pmatrix} \right) \frac{x^{c-(k-1)}}{(c-(k-1))!}  \\ 
=\left(\sum_{k=0}^{min(a,c+)} \frac{y^{a-k}}{(a-k)!}\begin{pmatrix} h-a-c+2(k-1) \\ (k-1)\end{pmatrix}\frac{x^{c-(k-1)}}{(c-(k-1))!}\right) + \left(\sum_{k=0}^{min(a,c+1)}\frac{y^{a-k}}{(a-k)!} \begin{pmatrix} h-a-c+2(k-1) \\ k \end{pmatrix} \frac{x^{c-(k-1)}}{(c-(k-1))!}\right)  \\
=\left(\sum_{k=1}^{min(a,c+1)} \frac{y^{a-k}}{(a-k)!}\begin{pmatrix} h-a-c+2(k-1) \\ (k-1)\end{pmatrix}\frac{x^{c-(k-1)}}{(c-(k-1))!}\right) + \left(\sum_{k=0}^{min(a,c+1)}\frac{y^{a-k}}{(a-k)!} \begin{pmatrix} h-a-c+2(k-1) \\ k \end{pmatrix} \frac{x^{c-(k-1)}}{(c-(k-1))!}\right)  \\
=\left(\sum_{k=0}^{min(a,c+1)-1} \frac{y^{a-1-k}}{(a-1-k)!}\begin{pmatrix} h-a-c+2k \\ k\end{pmatrix}\frac{x^{c-k}}{(c-k))!}\right) + \left(\sum_{k=0}^{min(a,c+1)}\frac{y^{a-k}}{(a-k)!} \begin{pmatrix} h-a-c+2(k-1) \\ k \end{pmatrix} \frac{x^{c-(k-1)}}{(c-(k-1))!}\right)  \\
=\dots$
Still I don´t know how to bring this together.
Thank you for helping me!

You do need to keep careful track of the terms for the induction to go smoothly. In addition to the listed steps my calculation needed the formula (proof is straightforward)
$$
k\binom n k=(n-k+1)\binom n {k-1}.\qquad(*)
$$
I abbreviate 
$$
y^{[k]}=\frac{y^k}{k!}
$$
and similarly with $x^{[k]}$. Things are a bit simpler if we declare $x^{[-1]}=y^{[-1]}=0$. So we have the obvious formula
$$
x\cdot x^{[k]}=\frac1{k+1}x^{[k+1]}.
$$
With all that the inductive step $c\to c+1$ proceeds as follows
$$
\begin{array}{lcl}
x^{[c+1]}y^{[a]}&=&\dfrac x{c+1}x^{[c]}y^{[a]}\\
&\overset{\text{ind.hyp.}}=
&\displaystyle\frac1{c+1}\sum_{k=0}^{\min(a,c)}xy^{[a-k]}\binom{h-a-c+2k}kx^{[c-k]}\\
&\overset{c=1}=&
\displaystyle\sum_{k=0}^{\min(a,c)}\frac1{c+1}
\left\{y^{[a-k]}x+y^{[a-k-1]}(h-a+k+1)\right\}\binom{h-a-c+2k}kx^{[c-k]}\\
&=&\displaystyle\sum_{k=0}^{\min(a,c)}\frac1{c+1}y^{[a-k]}x\binom{h-a-c+2k}kx^{[c-k]}\\
&+&\displaystyle\sum_{k=0}^{\min(a,c)}\frac1{c+1}y^{[a-k-1]}(h-a+k+1)\binom{h-a-c+2k}kx^{[c-k]}\\
&=&\displaystyle\sum_{k=0}^{\min(a,c)}\frac1{c+1}y^{[a-k]}\binom{h-2-a-c+2k}kx\cdot x^{[c-k]}\\
&+&\displaystyle\sum_{k=1}^{\min(a,c+1)}\frac1{c+1}y^{[a-k]}(h-a+k)\binom{h-a-c+2k-2}{k-1}x^{[c-k+1]}\\
&=&\displaystyle\sum_{k=0}^{\min(a,c)}y^{[a-k]}\frac{c+1-k}{c+1}\binom{h-2-a-c+2k}k x^{[c+1-k]}\\
&+&\displaystyle\sum_{k=1}^{\min(a,c+1)}y^{[a-k]}\frac{h-a+k}{c+1}\binom{h-a-c+2k-2}{k-1}x^{[c-k+1]}.\\
\end{array}
$$
Comparing this with what we want we see that it suffices (modulo the two terms at the ends that are easy to check) to verify the identity
$$
\frac{c-k+1}{c+1}\binom{h-2-a-c+2k}k+\frac{h-a+k}{c+1}\binom{h-a-c+2k-2}{k-1}=\binom{h-a-c-1+2k}k
$$
To do that I rewrote the fractional multipliers of the binomial coefficients in the form
$$
\frac{c-k+1}{c+1}=1-\frac k{c+1}
$$
and
$$
\frac{h-a+k}{c+1}=1+\frac{h-a-c+k-1}{c+1}.
$$
With the $1$s you use the Pascal triangle identity to get what you want. The extra terms cancel by identity $(*)$.
