how can i get a contradiction (number theory) i have some problems to get a contradiction:
problem: Does not exist $n > 0$ such that $a = p^{n}$. prove if $\log_{p}(a)$ is irrational (where $a \in \mathbb Z$ and $p$ prime) 
i try this:
suppose $\log_{p} (a)$ is rational 
then $\exists s,t$ such 
$\log_{p} (a) =\frac{s}{t}$ 
$\implies p^{\frac{s}{t}}$ = a 
$\implies \sqrt[t]{p^{s}}$ = a
$\implies p^{s} = a^{t}$ ( i think this is a contradiction)
if not how can i get a contradiction, can someone give me a hint 
thanks for reading me
 A: When you arrive at $p^s = a^t$, use the fundamental theorem of arithmetic to argue that the only prime factor of $a$ is $p$.
Hence $a$ must be an integer power of $p$, contradicting the non-existence of $n$ such that $a = p^n$.
A: Suppose ${\rm log}_p(a) = \frac{n}k,\ \gcd(n,k)=1.\,$ Then $\, a = p^{\large n/k} \Rightarrow\, a^{\large\color{#c00} k} = \color{#c00}p^{\large n},\,$ hence this theorem  implies that $\,\color{#c00}p\,$ is a $\color{#c00}k$'th power, $ $ so $\:k=1,\:$ so we conclude that $\,a = a^{\large k} = p^{\large n}.$
A: Um...... $a = p^n \iff \log_p a = n$. That's a definition.
So if $\log_a p$ is irrational then $\log_a p\ne n$ for any integer and $a\ne p^n$ for any integer.
(Of course, we have to assume (prove) that the definition of "$\log_p a$" makes any sense.  That is that for $a>0, a\ne 1, p > 0$ there is a unique $x\in \mathbb R$ so that $a^x =p$.  And for that we have to define/figure out what $a^x$ means.  .... But none of that is in the realm of number theorem, I think.)
