How can I prove that for $K\subset L$ - the Galois extension of degree $p^n$, where $p$ is prime, there exists a sequence $K=K_{0}\subset K_{1}\subset \cdots \subset K_{n}=L$ such that $[K_{i}:K_{i-1}]=p$, $i=1,\ldots,n$ and $K\subset K_i$ is the Galois extension either?
I know that for Galois extensions the degree of the extension is equal to the number of intermediate subfields, but it doesn't imply that there exists the sequence mentioned above.
I would be grateful for any help.
First of all, the degree of the extension in general can be different from the number of intermediate subfields. For example, an extension of prime degree $p>2$ has no nontrivial subextensions!
To answer your question, notice that it is enough to show that there is a $K_{n-1}$ such that $[K\colon K_{n-1}]=p$, because then you proceed by induction since $[L\colon K]=[L\colon K_{n-1}][K_{n-1}\colon K]$. The existence of such a $K_{n-1}$ is guaranteed by group theory: a group of order $p^n$ has nontrivial center, hence by Cauchy's theorem it has a subgroup $H$ of order $p$ which is a subgroup of the center and therefore is normal. Then $K_{n-1}$ is just the fixed field of $H$.