Help on proof of $\binom{n}{0}^2 + \binom{n}{1}^2 + ... + \binom{n}{n}^2 = \binom{2n}{n}$

Doubt 1: it isn´t an equation, but an identity (!!!): $${{a}^{n}}\cdot {{a}^{n}}\equiv {{\left( {{a}^{n}} \right)}^{2}}\equiv {{a}^{2n}},$$ or $${{a}^{n}}\cdot {{a}^{n}}\equiv {{a}^{n+n}}\equiv {{a}^{2n}}.$$Yes, Mathematics are indisputably important, and for many, also beautiful. But much more important than Mathematics and humility and gratitude.

In the brief indication that I have given you about your doubt number 1, there is the answer to all your doubts. However, it seems that my correction has bothered you.

Your algebraic development is correct and the only thing you need is to understand the validity of the arguments used. And all this is based on distinguishing between equation and identity: an equation is an equality that is verified only for certain values ​​of the variables, or for none, while an identity is an equality that is fulfilled for any value of the variables in its definition domain.

Since the equality $(1 + x) ^ n (1 + x) ^n = (1 + x) ^ {2n}$ is an identity, and two polynomials are identical (they take the same value for any value that is assigned to their variable) if and only if the coefficients of the terms of equal degree are identical, the coefficient of $x ^ n$ is the same in the development of $(1 + x) ^ n (1 + x) ^n$ than in that of $(1 + x) ^ {2n}$.

This simple reasoning is the idea underlying the validity of the argument used in the demonstration, that is, proving that the coefficient of $x ^ n$ is the same in the development of $(1 + x) ^ n (1 + x) ^n$ than in that of $(1 + x) ^ {2n}$.

Therefore, the answer given in my previous post was not trivial; It contained the key to solving all your doubts about the validity of your demonstration: identity is unconditional equality and in the case of the identity of polynomials, this unconditionality implies the equality of its coefficients in terms of the same degree. Keep in mind that a polynomial can be defined by omitting its variable, simply by the succession of its coefficients, so that two polynomials are identical if the sequences of their coefficients are equal.

Robert Shore (https://math.stackexchange.com/users/640080/robert-shore), Help on proof of $\binom{n}{0}^2 + \binom{n}{1}^2 + ... + \binom{n}{n}^2 = \binom{2n}{n}$, URL (version: 2019-04-17): https://math.stackexchange.com/q/3191796 gives you the same clarifications.