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For natural numbers, I would like to prove that if $m < n$, then $k^m < k^n$ for $k$ not $0$ or $1$. Induction seems viable, but I don't know which variable to induct on. Any suggestions?

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Use induction on $k$ (if you really want to prove this by induction); the object is to show this for all $k \in \mathbb{N}$ with $k > 1$. The base case is $k = 2$; you get more twos in $2^n$ than in $2^m$, so $2^m < 2^n$. For the inductive step you need to show $k^m < k^n \implies (k + 1)^m < (k + 1)^n$; maybe try the Binomial Theorem. Alternatively, try using Calculus knowledge; $f(x) = x^m$ and $g(x) = x^n$ are both increasing functions and $f(1) = g(1) = 1$; taking $m$ derivatives might shed some insight. –  user4689 Mar 2 '11 at 18:46
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Induction is viable, but not the most direct way. Because $k^n=k^m k^{n-m}$, you only need to show that $k^p>1$ for $k>1$ and $p>0$ –  leonbloy Mar 2 '11 at 19:22
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It would be very helpful if you could please elaborate on your earlier comment that you are "developing naturals set theoretically", so that we know what methods are available to you. The answer could vary widely depending on the foundational level that you are working at. If the question is from a book then a simple reference would suffice to convey that contextual information. –  Bill Dubuque Mar 2 '11 at 19:49

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Hint: Consider $f(x)=a^{x}$ and prove that it's monotonically increasing.

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