$p^i\mathbb {Z}_p$ is an open subgroup of $\mathbb {Z}_p$ of index $p^i$, where $\mathbb {Z}_p$ is the group of p-adic integers

Question

I can't seem to understand the logic behind the following quote:

For each $i$, $p^i\mathbb {Z}_p$ is the kernel of the projection $\pi_i:\mathbb {Z}_p\rightarrow\mathbb Z/p^i\mathbb Z$.
Thus, $p^i\mathbb {Z}_p$ is an open subgroup of $\mathbb {Z}_p$ of index $p^i$.

Why is the second part derived from the first?
I tried proving $p^i\mathbb {Z}_p$ is an open subgroup of $\mathbb {Z}_p$ the old fashion way, using open neighborhoods of some $x\in p^i\mathbb {Z}_p$, but got nowhere..

Note:
In my source, $\mathbb Z_p$ is treated as profinite, i.e as the inverse limit of $\mathbb {Z}/p^i$. Hence, $\mathbb Z_p$ is a subset of the product of $\mathbb {Z}/p^i$ and $\pi_i$ is a restriction of the projection mapping to that subset. Under these conditions, $\pi_i$ is continuous.

Answer 1

Since $\Bbb Z/p^i\Bbb Z$ has the discrete topology, $\{0\} $ is an open subset of $\Bbb Z/p^i\Bbb Z$, consequently, $\operatorname {Ker}\pi_i=\pi_i^{-1}\{0\}$ is open being the preimage of an open.

Answer 2

There is a simple trick that proves that $p^i \mathbb{Z}_p$ is open subgroup of $\mathbb{Z}_p$.

Just write the expression as follows: \begin{align} p^i \mathbb{Z}_p &= \{ x \in \mathbb{Q}_p: |x|_p \leq p^{-i} \} \ \text{implies closed} \\ &= \{x \in \mathbb{Q}_p: |x|_p <p^{-i+1} \} \ \text{implies open} \end{align}