# Prove $25^n>6^n$ using induction

This seem too simple that I cant even break this down..

Base case: For $n=1$, we have: LHS: $25^1=25$; RHS: $6^1$

So LHS$>$RHS, holds.

Inductive, hypothesis: Assume $25^k>6^k$ for some $n=k>=1$

Inductive step: We nee to show that $k+1$ holds, that is

$25^{k+1}>6^{k+1}$

from LHS:

$25^{k+1} = 25 * 25^k$

how do I continue from this, its already larger than $6^{k+1}$, however, how can I use what I already know?

• 25*25^k > 25*6^k Dec 6 '15 at 2:51

As you said, we know $25^{k + 1} = 25^{k} \cdot 25$. Now, $25^{k} \cdot 25 > 25^{k} \cdot 6$, right? And by assumption, since $25^{k} > 6^{k}$, we have $25^{k} \cdot 6 > 6^{k} \cdot 6 = 6^{k + 1}$.
So, everything I said above is just: $$25^{k + 1} = 25^{k}\cdot 25 > 25^{k} \cdot 6 >6^{k} \cdot 6 = 6^{k + 1}$$