Consider the extension $K_i$ of $\mathbb{Q}_p$ by adjoining $(p^i-1)^{th}$ roots of $\pi$ Let us consider the $p$-adic field $\mathbb{Q}_p$. Let $\pi$ be the uniformizer of $\mathbb{Q}_p$. Consider the extension $K_i$ of $\mathbb{Q}_p$ by adjoining  $(p^i-1)^{th}$ roots of $\pi$, i.e., $K_i=\mathbb{Q}_p(\sqrt[p^i-1]{\pi})$ such that $K_i \subseteq K_{i+1}$. Then take union $K=\bigcup_{i=1}^{\infty} K_i$. Then obviously $K$ is an extension of $\mathbb{Q}_p$, probably "infinite extension".
My question- 
What kind of extension $K$ is ? 
Is it algebraic extension extension ? 
Is it maximal unramified or maximal abelian extension ?
I know that for any algebraic extension $F$, it can be written as a union of finite subextensions i.e., $F=\bigcup_{i=1}^{\infty} K_i$ such that $[K_i: \mathbb{Q}_p]< \infty$ and $K_i \subseteq K_{i+1}$.
Help me with the above three questions. 
 A: For convenience, write $n_i=p^i -1$ (since you don't give your motivations, I can't see the reason for the explicit form of $n_i$). Introduce $K_i=\mathbf Q_p(\sqrt [n_i] \pi)$ and $F= \cup K_i$. Preliminary remark: there is no inclusion $K_i \subset K_{i+1}$ . Actually, the polynomial $X^{n_i} -\pi$ is Eisenstein, so $[K_i : \mathbf Q_p]=n_i$, but a priori $n_i$ does not divide $n_{i+1}$. So you can make sense only by considering the compositum $F$ of all the $K_i$'s. You ask 3 questions about $F$ :
1) Is $F/\mathbf Q_p$ an algebraic extension ? Sure, by construction it is an infinite algebraic extension.
2) Is it maximal unramified ? No, it isn't even unramified, since any $K_i$ is totally ramified (Eisenstein).
3) Is it maximal abelian ? No, by construction it is not even galois ! To get a galois extension over $\mathbf Q_p$, introduce the compositum $K$ of $F$ and of all the cyclotomic fields $\mathbf Q_p(\sqrt [n_i] 1)$. But $K$ will not be abelian since it has subextensions (all the $K_i$'s) which are not galois over $\mathbf Q_p$.                                                                                                                                                                                                                            
