I am trying to expand $\quad (x + 2)^3 $

I am actually not to sure what to do from here, the rules are confusing. To square something is simple, you just foil it. It is easy to memorize and execute. Here though I am not sure if I need to do it like multiplication where I take one $(x + 2)$ term and multiply by another or if I need to multiply all $(x + 2)$ terms by it. 

I want to treat it like how I would square it so I just square it and then I am left with the result and the $(x + 2)$ term. This is wrong and I do not know why. I get this

$$(x^2 + 4x + 4)(x + 2)$$

This is wrong and I am not sure why. So not I try the other way, multiplying everything by everything. This leaves me with

$$(x^2 + 4x + 4)(x^2 + 4x + 4)$$

Which is again wrong. I have exhausted all my options and nothing results in a correct answer and I am not sure why.


Question: Prove that a primitive root of $p^2$ is also a primitive root of $p^n$ for $n>1$.
Answer: Let us give a more elementary answer, while still using some binomial theorem. But we shall employ no more than a binomial lemma. It states that, for a prime $p$, and an integer $1\leq a\leq p-1$, we have $p\mid \binom{p}{a}$.
Now you know that $a$ is a primitive root of $p^2$, so the order of $a$ modulo $p^2$ is $p(p-1)$. And we know that $a^{p-1}=1+kp$ for some $k$. Then the assumption that $a$ is a primitive root of $p^2$ implies that $k$ is not divisible by $p$. Therefore $$a^{p^n(p-1)}=(a^{p-1})^{p^n}=1+kp^{n+1}+mp^{n+2}$$ for some $m$.(This is the place where we make use of the binomial lemma.) This directly tells you that $a$ is a primitive root of $p^n$ for every $n\ge1$.  


   More Details.  We elaborate upon two things: Firstly, we show the centered equation, and, secondly, we show how that implies the primitivity of $a$.  For the first, use the lemma to deduce that $(a^{p-1})^{p}=1+kp\times p+\text{terms of higher powers of $p$}$. Now, by induction, the result follows. For the second, just divide $(a^{p-1})^{p^r}$ by $p^n$ for $k=0,1,\ldots,n-1$, to see that, for lower powers of $a$ than $(a^{p-1})^{p^{n-1}}$, it cannot be congruent to $1$ modulo $p^n$. So $a$ is indeed a primitive root of $p^n$.  


If there is any error in the above proof, please inform me; if there is any ambiguity, please point it out, thanks.