Lie bracket in the universal enveloping lie algebra of $A_1$ 
Consider the universal enveloping algebra of $A_1$ with the standard basis $\{e,h,f\}$ where $[h,e]=2e, [h,f]=-2f, [e,f]=h$. I want to show that 
  $$[e,f^{\otimes n}]=n(n-1)f^{\otimes (n-1)}+n h\otimes f^{\otimes(n-1)}$$ for $n\in\mathbb{N}$.
Here we have defined $f^{\otimes 0}=1, f^{\otimes 1}=f, f^{\otimes 2}=f\otimes f, f^{\otimes 3}=f\otimes f\otimes f, ...$ and so on.

My idea is to do this by induction, however I seem to get stuck everytime.
 A: For a Lie algebra $\mathfrak{g}$, we have the following Leibniz product rule: $$\begin{align}[a,bc]&=a(bc)-(bc)a=(abc-bac)+(bac-bca)\\&=(ab-ba)\,c+b\,(ac-ca)=[a,b]\,c+b\,[a,c]\end{align}$$
for any $a,b,c\in \mathfrak{U}(\mathfrak{g})$.  We let $\mathfrak{g}:=\mathfrak{sl}_2$ with basis $\{e,h,f\}$ such that
$$[e,f]=h\,,\,\,[h,e]=+2\,e\,,\text{ and }[h,f]=-2\,f\,.$$
Note that
$$\mathfrak{U}(\mathfrak{g})=\text{span}\big\{e^ph^qf^r\,\big|\,p,q,r\in\mathbb{Z}_{\geq 0}\big\}\,.$$
From the Leibniz rule, we obtain
$$[e,f^n]=[e,f^{n-1}]\,f+f\,[e,f^{n-1}]$$
for every integer $n\geq 1$.  We can induct on $n$ to obtain the following binomial formula
$$[e,f^n]=\sum_{r=0}^{n-1}\,\binom{n-1}{r} \,f^rhf^{n-1-r}$$
for all positive integers $n$.  However, if we want to write $[e,f^n]$ in the basis $\big\{e^ph^qf^r\,\big|\,p,q,r\in\mathbb{Z}_{\geq 0}\big\}$ of $\mathfrak{U}(\mathfrak{g})$, we need to use the equality
$$fh=hf-[h,f]=hf-(-2f)=hf+2f\,.$$
Then, induction on $n$ yields
$$[e,f^n]=2^{n-1}\,hf^{n-1}+2^{n-1}(n-1)\,f^{n-1}\,.$$
This is different from the OP's assertion.
Similarly,
$$[e^m,f]=\sum_{r=0}^{m-1}\,\binom{m-1}{r}\,e^rhe^{m-1-r}=2^{m-1}\,e^{m-1}h-2^{m-1}(m-1)\,e^{m-1}$$
for every positive integer $m$.  I wonder what $[e^m,f^n]$ looks like for arbitrary  positive integers $m$ and $n$.  For example,
$$[e^2,f^2]=4ehf+8ef-2h^2-2h\,.$$
This looks to be quite a challenge.
Remark.  It is not recommended to use $\otimes$ as the multiplication in $\mathfrak{U}(\mathfrak{g})$.  When you write $a\otimes b$, people may interpret this as an element of $\mathfrak{U}(\mathfrak{g})\otimes \mathfrak{U}(\mathfrak{g})$.  
