# Thread: Proof by induction question

1. ## Proof by induction question

Hello guys

I'm having a problem with this real simple proof by induction example.

Proof is that $5^n>n^5$ for all integers $n>6$

1. anchor step $P(6)$: $5^6=15625=7776=6^5$

This is pretty obvious, ok so far

2. inductive step
Assume $P(n)$
$5^{n+1}=5\cdot5^n>5n^5$
Ok here is the problem. we get $5^{n+1}=5\cdot5^n$ from sub-in $n+1$ into $5^n$ and law of exponentials
But how do we get the other side $5n^5$ ???
$5n^5$ is then used in the last part of the proof:

Since $n\geq6$ and $P(n)$ is true

$(n+1)^5=n^5(\frac{n+1}{n})=n^5(1+\frac{1}{n}) \leq n^5(1+\frac{1}{6})^5=n^5(2.16...)<5n^5$

Together we have $5^{n+1}>(n+1)^5$
So we've shown that $P(n)\Rightarrow P(n+1)$
Thanks guys for the help

2. ## Re: Proof by induction question

Originally Posted by shakra
Hello guys

I'm having a problem with this real simple proof by induction example.

Proof is that $5^n>n^5$ for all integers $n>6$

This is pretty obvious, ok so far

Ok here is the problem. we get $5^{n+1}=5\cdot5^n$ from sub-in $n+1$ into $5^n$ and law of exponentials
But how do we get the other side $5n^5$ ???

Same comment as the other proof you posted. You are assuming that $5^n>n^5$ so under that assumption $5*5^n>5*n^5$

$5n^5$ is then used in the last part of the proof:

Thanks guys for the help
see comment in red

3. ## Re: Proof by induction question

Since when is 15 625 = 7776???

4. ## Re: Proof by induction question

I guess that was meant to be 15625 > 7776.