Is $\sum_{c \notin \mathbb Z_q}\psi(c)-\sum_{c \notin \mathbb Z_q}\psi(c^n)$ an integer?

Question

Let $n$ be a positive integer. Let $p$ be an odd prime and $q=p^k$. Let $c \in \mathbb Z_q$. Consider the additive character $\psi:\mathbb Z_q \rightarrow \mathbb C^{\times}$ that is defined as $\psi(m)=e^{2\pi im/p^k}$. Is the following always an integer? $$\sum_{c \notin \mathbb Z^{\times}_q}\psi(c)-\sum_{c \notin \mathbb Z^{\times}_q}\psi(c^n)$$ where $\mathbb Z^{\times}_q$ is the multiplicative group of $\mathbb Z_q$

Answer 1

Since $\psi$'s domain is precisely $\Bbb Z_q$, both sums are empty sums, conventionally taken to be $0$. Thus, you have have $0-0=0$, an integer.