In the ring $\mathbb{Z}_p$, $p$ is prime, $(a+b)^p=a^p+b^p$ proof?

Your argument seems perfectly fine to me!

The reason the question gives this hint is that the same result holds in any commutative ring of characteristic $p$, and you obviously can't use the same proof for rings of order $\neq p$.

So to prove the result in general, let $R$ be a commutative ring of characteristic $p$, and let $a, b \in R$. Then by the Binomial Theorem, we have $$ (a+b)^p = a^p + \binom{p}{1}a^{p-1}b + \binom{p}{2}a^{p-2}b^2 + \ldots + \binom{p}{p-1}ab^{p-1} + b^p. $$

Now, for $1 \leq k \leq p - 1$, we have $$ \binom{p}{k} = \frac{p!}{k!(p-k)!}, $$ and $p$ divides the numerator but not the denominator (since $p$ is prime), so $\binom{p}{k}$ is a multiple of $p$, and hence it is zero in $R$. Therefore, all the terms in our big sum are zero, except for the first and last ones, which gives us $$ (a+b)^p = a^p + b^p. $$

The Binomial Theorem says:

$(a + b)^{p} = a^{p} + \binom{p}{1}a^{p-1}b + \binom{p}{2}a^{p-2}b^{2} + ... + \binom{p}{p - 1}ab^{p-1} + b^{p}$

Where $\binom{n}{k}$, the binomial coefficients, $ = \frac{n!}{k!(n-k)!} = \frac{n(n-1)(n-2)...(n-k+1)}{k(k-1)(k-2)...(1)}$. (Intuitively, $\binom{n}{k}$ is defined as the number of ways to pick $k$ objects out a group of $n$, which is why it is read "$n$ choose $k$". This definition may also give you some insight into why the Binomial Theorem works.) Note that if $n$ is prime and $1\leq k \leq n-1$, then $n\vert \binom{n}{k}$ because there are no factors of $n$ in the denominator. Then, because all terms except $a^{p}$ and $b^{p}$ contain binomial coefficients divisible by $p$, we have $\boxed{(a + b)^{p} = a^{p} + b^{p}}$

You used Fermat's Little Theorem instead, which also works.

You should use binomial theorem because the result will be more general and argument more elementary. The proof will work for any commutative ring with characteristic $p$ instead of just $\mathbb{Z}_p$.

The key comes down to one fundamental property of prime number.

If $p$ is a prime number and $p | ab$, then $p|a$ or $p|b$.

For any integer $k$ with $1 \le k \le p-1$, the binomical coefficients $\binom{p}{k}$ are given by following expression.

$$\binom{p}{k} = \frac{p!}{k!(p-k)!}\quad\iff\quad\binom{p}{k} k! (p-k)! = p!$$

Since $p$ divides RHS, $p$ divides one of the factors on LHS. Since $k!$ and $(p-k)!$ are products of integers smaller than $p$. $p$ doesn't divide $k!$ nor $(p-k)!$. This forces $p$ divides $\binom{p}{k}$. In other words, $\frac1p \binom{p}{k}$ is an integer.

Apply binomial theorem to $(a+b)^p$, we get

$$(a+b)^p = \sum_{k=0}^p \binom{p}{k} a^k b^{p-k} = a^p + b^p + p \sum_{k=1}^{p-1} \left[\frac1p \binom{p}{k}\right]a^kb^{p-k} $$ In any commutative ring with characteristic $p$, multiply something by $p$ kills it.
This leaves us with $(a+b)^p = a^p + b^p$.

About the direction you are heading on, you can achieve what you want using following fact:

$\mathbb{Z}_p^{*} = \{ z \in \mathbb{Z}_p : z \ne 0 \}$, the non-zero elements of $\mathbb{Z}_p$, forms a group under multiplication.

For any $z \in \mathbb{Z}_p^{*}$, consider the sequence of elements $1, z, z^2, z^3, \cdots$. Since $\mathbb{Z}_p^{*}$ is finite, this sequence will repeat itself somewhere. From that, you can deduce for some positive integer $d$, $z^{d} = 1$. Furthermore, if $d$ is the smallest integer for $z^{d} = 1$, you can show the $d$ elements $1, z, z^2, \cdots, z^{d-1}$ forms a subgroup of $\mathbb{Z}^{*}$ with respect to multiplication.

By Lagrange theorem, $d$ divides $| \left|\mathbb{Z}_p^*\right| = p-1$. This implies $$z^{p-1} = \left(z^d\right)^{\frac{p-1}{d}} = 1^{\frac{p-1}{d}} = 1$$ Multiply $z$ by both side, you will find for any $z \in \mathbb{Z}_p$, one has $z^p = z$.

Please note that same argument works for any finite field. If $F$ is a finite field with $n$ elements, then all its elements are roots of the polynomial $z^n - z = 0$.