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My thinking is to let the $N = N_1^2$ (where $N_1$ is the N chosen for $a_n$). Is this correct? How to justify this?

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    you don't need to proceed in algebraic mode. Choosing $N = N_{1}$ is sufficient where $N_{1}$ is chosen for $a_{n}$ and $N$ is for $a_{n^{2}}$. Why? Because $N^{2} = N_{1}^{2} \geq N_{1}$. – 2017-02-08

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Rather, take $N = \lfloor \sqrt N_1 \rfloor + 1$ (then, for $n \ge N$, $n \ge \sqrt N_1$, so $n^2 > N$ and so $|a_{n^2} - L| < \epsilon$). Another way to do it would be using the more general fact that all the subsequences of a convergent (to $L$) sequence converge to $L$.

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    While your answer is correct, it does have the tone of algebra. Choosing $N = N_{1}$ suffices. – 2017-02-08
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    @ParamanandSingh I did that merely to tackle the problem in the OP's solution. – 2017-02-08
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    I don't personally prefer such algebraic manipulations in $\epsilon, \delta$ proofs. The continued focus on algebraic manipulations (a habit continued from primary and high school) is the reason students don't feel comfortable with such questions. This is also the reason OP asked this question. continued.. – 2017-02-08
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    We just need to note that $n^{2} > n$ and hence if $|a_{n} - L| < \epsilon$ for $n > N$ then obviously $n^{2} > n > N$ and hence $|a_{n^{2}} - L| < \epsilon$ so that the same $N$ works for $a_{n^{2}}$. – 2017-02-08
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    @ParamanandSingh I understand. Again, my answer was my way of telling the OP that they should've thought of "square root" instead of "square". – 2017-02-08