if $\displaystyle p|a^2$, then $\displaystyle p|a$.
This is true but how to prove it?
$\displaystyle p|a^2\rightarrow p|a*a$ now what?
If I wanted to generalize it, I could then say:
if $\displaystyle p|a^n$, then $\displaystyle p|a$.
$\displaystyle p|a^n\rightarrow p|a_1*a_2\dots *a_n\rightarrow p|a_1 \ \mbox{or} \ \dots \ p|a_n$ where $\displaystyle a=a_1=a_2=\dots=a_n$; hence, $\displaystyle p|a$
$\displaystyle P(n): \ p|a^n, \ n\geq 2$
$\displaystyle P(2): \ p|a^2\rightarrow p|a*a\rightarrow p|a \ \mbox{or} \ p|a$
$\displaystyle P(k): \ p|a^k$
$\displaystyle P(k+1): \ p|a^{k+1}$
Assume $\displaystyle P(k)$ is true.
$\displaystyle p|a^{k}$ multiple by a $\displaystyle a*p|a*a^k \rightarrow a*p|a^{k+1}\rightarrow p*(a*m)=a^{k+1}\rightarrow p|a^{k+1}\rightarrow p|a^k \ \mbox{or} \ p|a$
Suppose $\displaystyle p|a $ is false , then one must be coprime to the other , so there exists integers $\displaystyle x,y $ such that $\displaystyle ax + py = 1 $
$\displaystyle a^n x + p (a^{n-1}y) = a^{n-1} $
Let $\displaystyle a^n = pm $ because we have $\displaystyle p|a^n $
$\displaystyle pmx + pa^{n-1}y = a^{n-1}$
$\displaystyle p (mx + a^{n-1}y ) = a^{n-1} $ , we find that $\displaystyle a^{n-1} $ is the multiple of $\displaystyle p$ say $\displaystyle a^{n-1} = pm' $ , followed by $\displaystyle a^{n-1}x + p a^{n-2} y = a^{n-2} $ , $\displaystyle p(m'x + a^{n-2}y) = a^{n-2} $ so we have $\displaystyle a^{n-2} = pm'' ~ .... $ it eventually goes to $\displaystyle a = p m^{(n-1)} $ , a contradiction .
Nothing against simplependulum's proof, but if we write $\displaystyle a^{k+1}=a\cdot a^k$ then we can complete the induction step as was done in the original problem (first few posts of the thread).
Edit: Oh I guess the OP already did that. Well here's how I would write it then.
Induction step: Assume $\displaystyle (p | a^k \Rightarrow p | a)$.
We want to show that $\displaystyle (p | a^{k+1} \Rightarrow p | a)$.
$\displaystyle p | a^{k+1} \Leftrightarrow p | a\cdot a^k \Rightarrow (p|a \lor p|a^k) \Rightarrow p|a$. QED.
Edit 2: Sorry sorry, I've been illiterate. chiph588@ already posted basically the same thing except for the very last step. But maybe my typesetting will be easier for the OP to follow..
Edit 3 (!): By the way, it's not necessary to use base case $\displaystyle n=2$. We can use $\displaystyle n=1$ since $\displaystyle p|a^1 \Rightarrow p|a$.