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!
 A: 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 $(*)$.
