Units of Witt vectors Let $R$ be a perfect ring of characteristic $p$, $W(R)$ its Witt vectors. We know that any element $x\in W(R)$ has a unique Teichmuller expansion:
$$x=[c_0]+[c_1]p+[c_2]p^2+\cdots$$
here all $c_i$ are in $R$. I want to know whether the following assertion is true:

$x$ is invertible $\Leftrightarrow$ $c_0$ is invertible.

The "$\Rightarrow$" direction is trivial. But can we prove the inverse?
Any help is appreciated. Thanks!
 A: To expand on Jyrki Lahtonen's comment, we can remove the perfect assumption. For simplicity, let us still suppose that $R$ is a commutative ring of characteristic $p$. Any element $x\in W\left(R\right)$ can be uniquely written as:
\begin{equation*}
x=\left[c_{0}\right]+V\left(x'\right)\text{,}
\end{equation*}
where $V$ is the Verschiebung operator, and $x'\in W\left(R\right)$. In the setting of your question, you can define $x'=F\left(\sum_{u\in\mathbb{N}^{*}}p^{u-1}F^{-1}\left(\left[c_{u}\right]\right)\right)$ (because you assumed that $R$ is perfect, but we shall not).
As you said, it is clear that when $x$ is invertible, so is $c_{0}$. Conversely, as the Teichmüller operator is multiplicative, we have $\left[c_{0}\right]\left[{c_{0}}^{-1}\right]=1$, and in particular:
\begin{equation*}
x\left[{c_{0}}^{-1}\right]=\left[c_{0}\right]\left[{c_{0}}^{-1}\right]+V\left(x'\right)\left[{c_{0}}^{-1}\right]=1+V\left(x'F\left(\left[{c_{0}}^{-1}\right]\right)\right)\text{.}
\end{equation*}
So, as Jyrki Lahtonen wrote, we can just suppose without loss of generality that $c_{0}=1$.
Now, the series $\sum_{u\in\mathbb{N}}\left(-V\left(x'\right)\right)^{u}$ converges for the $V$-adic topology, because we have assumed that $R$ has characteristic $p$, so that $\left(-V\left(x'\right)\right)^{u}\in V^{u}\left(W\left(R\right)\right)$. And a direct computation yields:
\begin{equation*}
\left(1+V\left(x'\right)\right)\times\sum_{u\in\mathbb{N}}\left(-V\left(x'\right)\right)^{u}=1
\end{equation*}
