I want to find the p-adic expansion of $\frac{1}{p}$ and $\frac{1}{p^r}$ in the field $\mathbb{Q}_p$.
So, do I have to solve the congruences $px \equiv 1 \pmod {p^n}, p^r x \equiv 1 \pmod { p^n }, \forall n \in \mathbb{N} $, respectively?
But.. these congruences do not have solutions, right?
What else could we do, in order to find the p-adic expansion of $\frac{1}{p}$ and $\frac{1}{p^r}$ in $\mathbb{Q}_p$ ?
In short, $\frac{1}{p^n}$ is its own $p$-adic expansion in $\mathbb{Q}_p$.
The congruence $px\equiv 1\mod p^n$ is equivalent to solving $$px-p^ny=1$$ for $x,y\in\mathbb{Z}$. This is impossible because it would imply that $p\mid 1$.
The $p$-adic expansion of a number $n$ looks like a Laurent series in $p$ with coefficients in $\mathbb{Z}/(p\mathbb{Z})$. So we have $$\frac{1}{p}=1\cdot p^{-1}+0+0\cdot p+0\cdot p^2+\ldots$$
Another way of seeing it is that every non-zero element of $\mathbb{Q}_p$ can be represented uniquely as $p^{n}z$ where $n\in\mathbb{Z}$ and $z\in(\mathbb{Z}_p)^{\times}$. The unit $z$ can be expanded as a $p$-adic integer series and then shifted $n$ units to the right or left by $p^{n}$.
Proof:
First, I prove that it is possible to represent elements of $\mathbb{Q}_p$ in the form mentioned above. Let $0\neq q\in\mathbb{Q}_p$. Now define $z:=q\cdot\vert q\vert_p$. Calculating the norm of $z$ shows that it is a unit in $\mathbb{Z}_p$: $$\vert z\vert_p=\left\vert q\vert q\vert_p\right\vert_p=\left\vert qp^{-\nu_p(q)}\right\vert_p=\vert q\vert_p\cdot \left\vert p^{-\nu_p(q)}\right\vert_p=p^{-\nu_p(q)}\cdot p^{\nu_p(q)}=1.$$ This gives $q=\vert q\vert_p^{-1}z$. Where $\vert\cdot\vert_p$ is the $p$-adic norm and $\nu_p(\cdot)$ is the $p$-adic valuation.
Second, I prove that there is only one such way to represent a given $p$-adic number. Let $p^nx=p^my$ with $m,n\in\mathbb{Z}$ and $x,y$ units in $\mathbb{Z}_p$. Then $p^{n-m}=yx^{-1}$. The right hand side shows that $p^{n-m}$ is a unit, so $m=n$ and $1=yx^{-1}$. Thus $x=y$ as well.
Applying this to your question, observe that $p^{-n}=p^{-n}\cdot 1$ is already of this form and that the $p$-adic expansion of $1$ is $1+0\cdot p+0\cdot p^2+\ldots$
If you would like to have a better grasp of this topic, I highly recommend getting a copy of $p$-adic Numbers: An Introduction. There is also a good amount of content on wikipedia.
An analogy
In $\mathbb{C}$, every non-zero element can be represented in polar form $z=re^{i\theta}$. Where $\vert z\vert=r$ and $e^{i\theta}$ is an element of the unit circle. The "unit circle" in $\mathbb{Q}_p$ is in some sense the units of $\mathbb{Z}_p$. So writing $q=p^{\nu_p(q)}u$ (where $u$ is a unit of $\mathbb{Z}_p$) is analogous to having a polar form for $p$-adic numbers.