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The question from the Koblitz p-adic numbers book is stated as follows:

Use the discussion of $n^s$ in section 2 to ccompute the following through the $p^4$-place:

(i) $11^\frac{1}{601}$ in $\mathbb{Q}_5$

(ii) $\sqrt{1/10}$ in $\mathbb{Q}_3$

(iii) $(-6)^{2+4\times7+3\times7^2+7^3+...}$

Section 2 discusses extending $f(s)=n^s$ to all p-adic integers. I'm currently stuck on the first part. What I have done is write out the p-adic expansion of 11, and find the p-adic expansion of 1/601 out to $p^4$. I recognize that 1/601 is in $\mathbb{Z}_5$, but I'm failing to understand what the expression means.

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    Yes, what did you compute for $1/601$ in $\mathbb{Q}_5$? Did you see [this question](http://math.stackexchange.com/questions/1186967/method-of-finding-a-p-adic-expansion-to-a-rational-number)? – 2017-02-17
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    I have seen that question. Out to $p^4$, I got $1+5^2$. Past that, I get $(a_i)=\left\{1,0,1,0,0,0,4,4,3,4,4,4,0,0,1,0,0,0,...\right\}$, which is hopefully correct. So far, I have written, for out to $p^4$, $11^{1/601}$ as $(1+2\times 5)^{1+5^2+...}=(1+5^2)*(1+5^2)^{5^2}$, thinking that $11^{1/601}=\lim_{n\to\infty}(11^{\alpha_n})$ where $1/601=\alpha=(\alpha_n)$ and that considering anything higher than $\alpha_4$ would contribute terms past $p^4$. Then I expanded $(1+5^2)^{5^2}$ out to the appropriate terms and got $1+2\times 5+4\times 5^2+3\times 5^3+2\times 5^4$ out the $p^4$. – 2017-02-17
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    But that doesn't seem right, so I'm unsure of how to proceed. For the second part of the question, I just solved for the equation $\alpha^2-(1/10)=0$ in $\mathbb{Z}/p^5\mathbb{Z}$. – 2017-02-17
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    Yes, $1/601=1+5^2$. Then what is $11^{1+5^2}$? – 2017-02-18
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    I think I understand now. Since $11\in 1+5\mathbb{Z}_5$ is a unit and 1/601 is in $\mathbb{Z}_5$, $$11^{1/601}=(1+2\times5)^{1/601}=\sum_{n=0}^{1/601}\binom{1/601}{n}(2\times5)^n$$ Just need to work out $\binom{n}{k}$ for a p-adic integer. Thanks – 2017-02-20

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