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I need to show that the equation $ x^2 = 2$ has a solution in $\mathbb{Z}_7$, where

$$\mathbb Z_7 = \lim_{\longleftarrow} A_n \space, (A_n = \mathbb{Z}/7^n \mathbb{Z})$$

Now I know that $\mathbb Z_7$ can also be represented in the form of power series as $$\sum_{n=0}^{\infty} a_n 7^n $$ for $0 \leq a_n \leq 6$. How do I use this information to solve my problem? Is there any other way to do this?

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    $3^2=9\equiv 2 \pmod 7$ Isn't that proof enough? – 2017-01-26
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    @AritraDas $\Bbb{Z}_7$ is referring to the $7$-adic integers here, not the integers modulo $7$. – 2017-01-26

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Are you aware of Hensel's Lemma? If so, just consider the polynomial $f(x)=x^2-2$, check that there is some $a\in\Bbb{Z}/7\Bbb{Z}$ such that $f(a)=0$ but $f'(a)\neq0$, and apply Hensel's Lemma.

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    You are correct. Check my edit, that was a typo. – 2017-01-26
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    Hensel's lemma works perfectly. Just out of curiosity, can this be done with involving this power series? – 2017-01-26
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    @Dark_Knight Possibly, but not that I know of. Hensel's Lemma is really your number one important tool for anything related to $p$-adic roots of polynomials. – 2017-01-26
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    Got it! Thanks! – 2017-01-26
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    @Dark_Knight You can prove by induction that there are $a_0,\dots,a_k$ such that $\sum_{n=0}^ka_n7^n$ is a square root of $2$ modulo $7^{n+1}$ and then take these as the coefficients of the power series. That would be just repeating the proof of Hensel's lemma though. – 2017-01-28
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Usually things can get clearer when considered from a general point of view. I'll try to do this by rendering algebraic your approach using power series. Let $K$ be a local field (for simplicity, you can just take $K=\mathbf Q_p$), with finite residual field $k$ (here $k=\mathbf F_p$), $A$ the ring of integers of $K$ (here $A=\mathbf Z_p$), $U$ the multiplivative group of invertible elements. Problem: for a given integer $m\ge 1$, determine the elements $x\in (K^*)^{m}$.

Let $v$ be the additive valuation of $K$, $\pi$ an uniformizer of $K$, i.e. $v(\pi)=1$, and $P=\pi A$, the maximal ideal of $A$. Using $v$, obviously $K^* \cong \mathbf Z \times U$. To further unwind $U$, introduce the filtration of subgroups $U_n = 1+P^n$, $n\geq1$, and consider the successive quotients $U_n /U_{n+1}$. Then (see e.g. chapter 1 of Cassels-Fröhlich's book):

1) $U/U_1 \cong k^*$. Since $k^*$ has order prime to $p$, Hensel's lemma implies that it can be lifted to $U$, i.e. $U$ contains a (maximal) subgroup $W$ of roots of unity of order prime to $p$. Then $U \cong W \times U_1$ because $W$ has order prime to $p$ and $U_1$ is a $\mathbf Z_p$-module

2) $U_n /U_{n+1} \cong P^n/P^{n+1} \cong (k, +)$

You'll convince yourself easily that the above results are the algebraic translation of the power series expansion in $K^*$. Let us apply this to the $m$-th power map $\mu_m :K^* \to K^*$:

3) If $m$ is prime to $p$, $\mu_m$ restricts to an automorphism of $U_n$ for any $n\geq1$

4) If $m=p$, then $\mu_p(U_n) \subset U_{n+1}$

5) If $m=p$ and $n$ is stricly larger than $e/(p-1)$ , where $e$ is the ramification index of $K/\mathbf Q_p$, then $\mu_p$ induces an isomorphism $U_n \cong U_{n+e}$

These are, I think, all the known general properties of $\mu_m$. Of course, your original problem is much simpler and is solved right from the start (property 1) above, or Hensel's lemma, as in @Alex Mathers answer).