If $x \in \mathbb{Z}/p^n\mathbb{Z}$, then $x = \mbox{unit} \times p^e$?

Let $R = \mathbb{Z}/p^n\mathbb{Z}$ where $p$ is a prime, and $n \ge 1.$ Let $\mathcal{U}(R)$ denote the units of $R.$ Is it possible to write any element $x \in R$ as $$x = up^e$$ where $u \in \mathcal{U}(R)$ and $0 \le e \le n$? Under what conditions does this factorization exist?

I am reading a paper, and it seems they assume they can always find this factorization.

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In fact any element $x \in \mathbb{Z}/p^n\mathbb{Z}$ has a unique "Taylor series" (or "decimal expansion") $x = c_0 + c_1 p + c_2 p^2 + ... + c^{n-1} p^{n-1}$ where $c_i \in \{ 0, 1, 2, ... p-1 \}$ (exercise). –  Qiaochu Yuan Sep 20 '11 at 5:46

Let $\tilde{x}$ be the integer with $0 \leq \tilde{x} < p^n$ which reduces to $x$ modulo $p^n$. I leave the case $\tilde{x} = 0$ to you. Otherwise we may write $\tilde{x} = p^e \tilde{u}$ with $\operatorname{gcd}(p,\tilde{u}) = 1$ and $0 \leq e \leq n$.
Then reducing modulo $p^n$ gives $x = p^e u$, where $u = \tilde{u} \pmod{p^n}$. Since $u$ is prime to $p$, it is a unit in $\mathbb{Z}/p^n \mathbb{Z}$.