Ring, nilpotent, and Binomial theorem Assume that R is a ring such that a prime number p is invertible in R. Show
that for any nilpotent element ε ∈ R the equation $x^p = 1 + ε$ has a solution in R.
So what I need to do is to show that $x =(1+ \epsilon)^{\frac1p} $ belongs to the ring .we know that $\epsilon$ is an nilpotent lets say of order n.
I open up the expression on the right and got
 $$x=1+ \frac1p \epsilon+\frac{\frac1p(\frac1p -1)}{2}\epsilon^2 + ...+ \frac{\frac1p(\frac1p-1)(\frac1p-2)*...*(\frac1p-n)}{(n-1)!}\epsilon^{n-1} + (...)*\epsilon^n +0 $$ p is invertible hence $\frac1p$ is on the ring, my problem is to understand how the other exppresion belongs to the ring .
I got to $\frac1{p^n}*\frac{(1-p)(1-2p)*..*(1-np)}{(n-1)!}$ and I cant see how the multiplication on the up is croosed out with the the factorial . why this thing belong to the ring and gives me my solution ?
ring is commutative with a unit
 A: Let $R$ be a commutative ring with unity, and suppose $p$ is a positive integer (not necessarily prime) such that $p$ is a unit in $R$.

Claim:$\;$If $\epsilon \in R$ is nilpotent, then $\;x^p = 1 + \epsilon,\;$for some $x \in R$.

Proof:

Suppose $\epsilon \in R$ is nilpotent of order $n$.

For $1 \le k \le n$, let $I_k = (\epsilon^k)$ be the principal ideal of $R$ generated by $\epsilon^k$.

Let $a_0=1$.

Claim:$\;\,$There exist $a_1,...,a_{n-1} \in R$ such that
$${\left(\sum_{i=0}^{k-1}\;a_i\epsilon^i\right)}^{\!p} \equiv 1+\epsilon\;\,(\text{mod}\;I_k)$$
for all $k$ with $1 \le k \le n$.

Proceed by induction on $k$.

For $k=1$, the claim reduces to
$$1  \equiv 1+\epsilon\;\bigl(\text{mod}\;(\epsilon)\bigr)$$
which is clearly true.

Next, assume the claim has been established for some fixed $k \ge 1$.

\begin{align*}
\text{Then}\;\;&{\left(\sum_{i=0}^{k-1}\;a_i\epsilon^i\right)}^{\!p} \equiv 1+\epsilon\;\,(\text{mod}\;I_k)\\[4pt]
\implies\;&{\left(\sum_{i=0}^{k-1}\;a_i\epsilon^i\right)}^{\!p} = 1+\epsilon + r\epsilon^k\,\;\text{for some $r \in R$}\\[4pt]
&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\text{Then, letting $a_k = -{\small{\frac{r}{p}}}$, we get}\\[4pt]
&{\left(\sum_{i=0}^{k}\;a_i\epsilon^i\right)}^{\!p} 
\,(\text{mod}\;I_{k+1})\\[4pt]
\equiv\;
&\left({\left(\sum_{i=0}^{k-1}\;a_i\epsilon^i\right)}+a_k\epsilon^k\right)^{\!p} 
 \,(\text{mod}\;I_{k+1})\\[4pt]
 \equiv\;
 &{\left(\sum_{i=0}^{k-1}\;a_i\epsilon^i\right)}^{\!p}+p{\left(\sum_{i=0}^{k-1}\;a_i\epsilon^i\right)}^{\!p}\!\left(a_k\epsilon^k\right)\;\,
 (\text{mod}\;I_{k+1})\\[4pt]
 \equiv\;
 &\left(1+\epsilon + r\epsilon^k\right)+\left(pa_0a_k\epsilon^k\right)\;
 (\text{mod}\;I_{k+1})\\[4pt]
 \equiv\;
 &1+\epsilon + \epsilon^k\left(r+pa_0a_k\right)\;
 (\text{mod}\;I_{k+1})\\[4pt]
  \equiv\;
  &1+\epsilon + \epsilon^k(0)\;\,
 (\text{mod}\;I_{k+1})\;\;\;\text{[since $a_0=1$ and $a_k = -{\small{\frac{r}{p}}}$]}
 \\[4pt]
 \equiv\;&1+\epsilon
\;\,(\text{mod}\;I_{k+1})\\[4pt]
\end{align*}
which completes the induction.

Then, letting $k=n$, we get
$$
{\left(\sum_{i=0}^{n-1}\;a_i\epsilon^i\right)}^{\!p} \equiv 1+\epsilon\;\,(\text{mod}\;I_n)
\qquad\qquad\qquad\qquad\;\;
$$
hence, noting that $I_n=0$, it follows that
$$
x = \sum_{i=0}^{n-1}\;a_i\epsilon^i
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\;\;
$$
satisfies the equation $x^p = 1 + \epsilon$.

This completes the proof.
