How do I prove by mathematical induction that $n!

Question

Suppose $n$ is equal or bigger than $3$. It's obviously true for $n=3$ that $n!<n^{n-1}$. To show more generally that

$$k! < k^{k-1} \text{ for some } k,$$

is it as simple as saying

$$(k+1)! = (k+1)k! \implies (k+1)!\lt k^k\:\:?$$

I'm new to induction so I am not sure I got it right. Thanks.

Sorry - do you want to show that $n!<n^n-1$ or $n!<n^{n-1}$? You have it both ways in the question. — Shane, Nov 05 '14 at 15:08
I believe it's cleared up now, it's n^(n-1) (I don't know how to format, I apologize for that) — John Doe, Nov 05 '14 at 15:11
Do you really need induction? $n! = 2\cdot3\cdots n<n\cdot n \cdots n = n^{n-1}$ — lhf, Nov 05 '14 at 15:13
Well it's obviously true, the exercise requires me to do it by induction, though... — John Doe, Nov 05 '14 at 15:14

score 4 · Accepted Answer · answered Nov 05 '14 at 15:14

4

Showing that it's true for $3$ $$3^2\gt3!$$ Assume that it's true for $k$ $$k^{k-1}\gt k!\tag{1}$$ we can see clearly that $$(k+1)^{k-1}\gt k^{k-1}$$ $$\implies (k+1)^{k-1}\gt k!$$ Multiplying Both side by $(k+1)$ $$(k+1)(k+1)^{k-1}\gt (k+1)k!$$

$$(k+1)^{k}\gt (k+1)!$$ if it's true for $k$ then it's true for $k+1$ and then by Induction Hypothesis....

answered Nov 05 '14 at 15:14

That's it. Thanks for your help! – John Doe Nov 05 '14 at 15:19
Done. I asked the question in a bit of a rush... didn't get to figure out the specifics of how the site works. – John Doe Nov 05 '14 at 15:22

How do I prove by mathematical induction that $n!

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