Let $\mathbb{Z}_p$ denotes the ring of $p$-adic integers, i.e., $\mathbb{Z}_p:= \varprojlim \mathbb{Z}/p^{n}\mathbb{Z}$. Then consider the projection map $\pi_{n}: \mathbb{Z}_p \to \mathbb{Z}/p^{n}\mathbb{Z}$. Then we know that $\pi_{n}$ is a ring map. I want to show that $\operatorname{ker} \pi_{n}=p^{n}\mathbb{Z}_p$.
It is clear that $p^{n}\mathbb{Z}_p \subset \operatorname{ker}\pi_{n}$. I am stuck at the converse. For $(x_r) \in \operatorname{ker}{\pi_n}$ we've $p^n \mid x_n$, and since $p^n \mid x_{n+1}-x_n$, $p^n \mid x_{n+1}$. Thus for all $i \in \mathbb{N}$ and $i\geq n$ there exists integers $y_{i}$ such that $x_i=p^{n}y_i$, so that we obtain $(x_r)=p^{n}(y_r)$ But I'm unable to show that $(y_r) \in \mathbb{Z}_p$, i.e., $(y_r)$ is coherent. I need some help to complete it. Thanks.
I would just write things out explicitly. We know that $x_n=0$. If we look at $x_{n+r} = \sum_{i=0}^{n+r-1}a_ip^i$, it restricts to $0$ mod $p^n$, hence $$x_{n+r} = \sum_{i=n}^{n+r-1}a_ip^i = p^n \sum_{i=0}^{r-1}a_{i+n}p^i = p^ny_r,$$ where $y_r = \sum_{i=0}^{r-1}a_{i+n}p^i \in \mathbb{Z}/p^r\mathbb{Z}$. As Phicar hinted, the coherence of $(y_r)$ now simply follows from the coherence of $(x_r)$. Explicitly, consider $x_{n+k} = \sum_{i=0}^{n+k-1}a_ip^i$ and $x_{n+j} = \sum_{i=0}^{n+j-1}b_ip^i$, where $k \geq j$. Then $x_{n+k} = x_{n+j}$ mod $p^{n+j}$, i.e. $p^n\sum_{i=0}^{j-1}a_{i+n}p^i = p^n\sum_{i=0}^{j-1}b_{i+n}p^i$. Therefore, $$y_j = \sum_{i=0}^{j-1}b_{i+n}p^i = \sum_{i=0}^{j-1}a_{i+n}p^i \implies y_k \text{ mod } p^j = \sum_{i=0}^{k-1}a_{i+n}p^i \text{ mod } p^j = \sum_{i=0}^{j-1}a_{i+n}p^i = y_j.$$