Proving that the first Weyl Algebra is simple The first Weyl algebra over the complex numbers $\mathbb{C}[x]$ is defined to be the set  $A_1 = {\{\sum_{i = 0}^{n} f_i(x) \delta^i : f_i(x) \in \mathbb{C}[x] }\}$. So it is the set of all linear differential operators with polynomial coefficients. $\delta$ is an operator meaning differentiation, and, $x$ is an operator meaning multiplication by $x$. As operators, the following relation holds; $(\delta \cdot x) = (1 + x \cdot \delta)$. 
Now, I've been reading a proof of the simplicity of the above algebra, but I don't understand it.
Let $I$ be a non-zero ideal of $A_1$. Choose a non-zero element $a \in I$. Hence, we may write $a = \sum_{i = 0}^{n} f_i(x)\delta^i$. Choose an $a$ such that $n$ is minimal. Then,the following is clearly in $I$ $$ -xa +ax  = \sum_{i=0}^{n} f_i(x)(-x\delta^i + \delta^ix) = \sum_{i=o}^nf_i(x)(i\delta^{i-1}) = nf_n(x)\delta^{n-1} + ... \neq 0 - (1)$$ 
This contradicts the minimality of $n$ and hence $n = 0$. Hence, $a = \sum_{j=0}^m\lambda_ix^i$, $\lambda_i \in \mathbb{C}$, with $\lambda_m \neq 0$.
Now, note that the following expression is in $I$ ; $$\delta a - a \delta = \sum_{j = 0}^m \lambda_i i x^{i-1} - (2)$$
Repeating this $m$ times we see that $\lambda_m(m!) \in I$, thus $1 \in I$, thus $I = A_1$ and we're done.
I don't understand how $(1)$ works? How are they getting these expressions? Does $ax$ mean applying the linear transformation $a$ to the polynomial $x$, or is it multiplication by $x$?
Also, going from $\sum_{i=0}^{n} f_i(x)(-x\delta^i + \delta^ix)$ to $\sum_{i=o}^nf_i(x)(i\delta^{i-1})$ seems like the $\delta^i$ has been differentiated by the $x$? How does this work?
How is it clear that the expression is nonzero? How is the minimality of $n$ contradicted here?
Also, in $(2)$, again, I'm unsure of how they've got that expression?
Sorry for all of the questions, I'm just really struggling to grasp this proof!
 A: For any ring, define the commutator by
$$[A,B]\stackrel{\text{def}}{=}AB-BA\text{.}$$
In any ring, this map is bilinear over the center and satisfies the product rule
$$\begin{align}[D,FG]&=[D,F]G+F[D,G]\\
[AB,X]&=A[B,X]+[A,X]B\text{.}
\end{align}$$
Now, the defining relation for the Weyl algebra is
$$[\delta,x]=1\text{.}$$
From this relation and the product rules above, we may prove that, for any univariate polynomials $f,a$, we have
$$\begin{align}
[\delta,f(x)]&=f'(x) &[a(\delta),x]&=a'(\delta)\\
[\delta,a(\delta)]&=0 &[f(x),x]&=0\text{.}
\end{align}$$
And now for the essence of the proof. Write $\mathsf{X},\mathsf{\Delta}$ for the $\mathbb{C}$-linear maps on $A_1$
$$\begin{align}\mathsf{X}a&\stackrel{\text{def}}{=}[a,x]\\
\mathsf{\Delta}a&\stackrel{\text{def}}{=}[\delta,a]\text{.}
\end{align}$$
Any nonzero element $a$ of $A_1$ may be written as
$$a=\sum_{i=0}^nf_i(x)\delta^i$$
where
$$f_n(x)=\sum_{j=0}^m\lambda_jx^j$$
and where $\lambda_m\neq 0$. ($\lambda_m\neq 0$ guarantees that $n$ is the order of $a$ and that $m$ is the degree of $f$. That every nonzero element of $A_1$ has a well-defined order basically corresponds to the minimality and non-zeroness part of your exposition.)
Then from the identities for the commutator above we have
$$\tfrac{1}{m!n!\lambda_m}\mathsf{\Delta}^m\mathsf{X}^na=1\text{;}$$
Thus every two-sided ideal containing a nonzero element $a$ contains $1$.
