How tounderstand the proof to show if a real number $w>0$ and a real number $b>1$, $b^w>1$? Claim: If there are two real numbers $w>0$ and $b>1$, then $b^w>1$. 
One proof is that for any rational number $0<r<w$ and $r=m/n$ with $m,n\in\mathbb Z$ and $n\ne0$, $(b^m)^{1/n}>1$. To show $b^m>1$, you can use induction, and then use $1^n=1<b^m$ to finally show that $b^r>1$. So the proof is done. 
I do not think this proof is complete. If $w$ is also a rational number, everything is fine. But if $w$ is an irrational number, I think the proof is incomplete, but I am not sure how to proceed. 
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Please do not use the continuity or more advanced techniques. Assume you just finished the first chapter of Rudin's Principles of Mathematical Analysis. 

Don't need continuity but need concept of convergence.
First realize that you actually don't have any frigging idea what $b^x $ actually means.  It's a bit of a shock to me but... it's true.  We have no definition that makes any sense.
But that's okay.  This is a perfect opportunity to define it.  Now that we've read chapter 1 of Rudin we actually know what the real numbers are which we didn't before.
Real numbers are the extension of the rationals with the least upper bound property.  So for all real $x $, there is one (many) sequences of rational {$q_n $} such that $q_n\rightarrow x $.  
So step 1.  Show that if $q_n <q_{n+1}<... $ converges to $x $ then $b^{q_n}<b^{q_n}<...$ and is bounded above.  I won't do that be its pretty simple.
So simply define $b^x = sup b^{q_n} $.
Step 2.  It's pretty obvious that if $b^{q_n} >1$ then $sup b^{q_n} $ is too.
See Rudin exercise 1.6 for more details.  
