Proving by induction $4^n < 5^n$ where $n > 0$

I have done a few normal inductions (like divisibility and sums) and know a little bit about strong induction. I can't find any problem similar to this one and am struggling to understand any intuition that would make proving this statement any easier.

Here is how far I have gotten (not very much):

Base Case:

$4^1 < 5^1$

$4^2 < 5^2$

$16 < 25$

Induction:

$4^k < 5^k$

$4^{k+1} < 5^{k+1} = 4\times4^k < 5\times5^k$

I have no clue where to go or how to force the k+1 equation to somehow equate to the k equation. Sorry if this has been asked before. Thank you.

Solution 1:

For the base case, directly checking the inequality $$ 4^1 < 5^1 $$ is enough.

Now, if for some positive integer $k$ we have $$ 4^{k} < 5^{k}, \quad \mbox{(*)} $$ we have to infer from it $$ \mbox{the desired inequality } 4^{k+1} < 5^{k+1}. $$

To make this inference, write $$ \begin{array}{llll} & 4^{k+1} & \\ = & 4^{k} & 4^1 \\ < & 4^{k} & 5^1 \quad \mbox{(by the base case)}\\ < & 5^{k} & 5^1 \quad \mbox{(by (*))}\\ = & 5^{k+1}. \end{array} $$

(So, basically, the task is to construct a chain of inequalities that leads to the desired result and where each inequality follows from something that's already been proved.)