I'm aware of several approaches to proving that for natural numbers if $n^2$ is even then $n$ must be even. I would appreciate some feedback on my attempt using the Fundamental Theorem of Arithmetic. Is the proof correct? Can it be improved in any way?
To prove: $\forall n \in \mathbb{N}, n^2$ is even $\implies$ n is even.
From the Fundamental Theorem of Arithmetic, we have
$$1 \lt n^2 = p_1^2p_2^2\dots$$ for some primes $p_i$.
If $n^2$ is even, then $p_{1} = 2$. Thus, taking the square root, we have
$$n = 2p_2p_3...$$
Since this is a multiple of $2$, we can conclude that $n$ is even. $\blacksquare$
It is correct in structure, but there is one line where the detail is not fully correct. For example, the step where you say $n^2 > 1 = p_{1}^{2}p_{2}^{2}..$ for some primes $p$. This should be broken down like:
1) Since $n^2$ is even, it is divisible by 2. Namely $n^2 > 1$
2) Using fundamental theorem of arithmetic, $n$ has a prime factorization $n = p_{1}^{k_1}...p_{m}^{k_m}$ for $k_i \geq 1$, $p_i$ primes. Thus $n^2$ has prime factorization $n^2 = p_{1}^{2k_1}...p_{m}^{2k_m}$
From there, the rest of your argument can be applied (without loss of generality, $p_1 = 2$, etc etc..)