In the theorem is it necessary for ring $R$ to be commutative? According to the statement of theorem that a commutative ring $R$ with prime characteristic $p$ satisfies 
$$\begin{align}
(a+b)^{p^n} = a^{p^n} + b^{p^n}
\end{align}$$
$$\begin{align}
(a-b)^{p^n} = a^{p^n} - b^{p^n}
\end{align}$$
Where a,b $\in$ $R$, n  $\geq 0$ and n $\in \mathbb{Z}$
I have thoroughly checked the proof many times but I don't understand why the condition of commutativity is important.

The commutative property is key in proving (for example, by induction) the Binomial Formula for commutative rings $A$ with $1$:
$$ (a+b)^n = \sum_{k=0}^n {n \choose k} a^k b^{n-k} $$
for all $a,b\in A$.
In prime characteristic $p$, we can then use $p | {p^n \choose k}$, $k<p^n$ and $k>0$. 
