This is a problem from Discrete Mathematics and its Applications

  1. Let $P(n)$ be the statement that $n!<n^n$, where $n$ is an integer greater than $1$.

$\quad(a)$ What is the statement $P(2)$?
$\quad(b)$ Show that $P(2)$ is true, completing the basis step of the proof.
$\quad(c)$ What is the inductive hypothesis?
$\quad(d)$ What do you need to prove in the inductive step?
$\quad(e)$ Complete the inductive step.
$\quad(f)$ Explain why these steps show that this inequality is true whenever $n$ is an integer greater than $1$.

I am currently on part e, completing the inductive step.
Here is my work so far,

I was able to show that the basic step, $P(2)$ is true because $2! < 2^2$ or $2 < 4$
Now I am trying to show the inductive step, or $P(k)\to P(k+1)$
Assuming $P(k)$, $k! <k^k$, show $(k+1)! < (k+1)^{k+1}$
To get $(k+1)!$ on both sides, I multiplied both sides by $k +1$ to get $$(k+1)! < k^k(k+1)$$ or $$(k+1)! < k^{k + 1} + k^k$$

How can I get this expression, $k^{k + 1} + k^k$ to equate $(k+1)^{k+1}$?

  • $\begingroup$ Should not have expanded. We have $(k^k)(k+1)\lt (k+1)^k(k+1)=(k+1)^{k+1}$. $\endgroup$ – André Nicolas Feb 25 '15 at 1:47
  • $\begingroup$ You don't need to "equate", it is sufficient to establish that $k^{k+1} + k^k < (k+1)^{k+1}$ $\endgroup$ – DanielV Feb 25 '15 at 1:49
  • $\begingroup$ @DanielV wait how did you turn that screenshot to text content? $\endgroup$ – committedandroider Feb 25 '15 at 1:52
  • $\begingroup$ @committedandroider Mr. Scott did that (and well), and to see how he did it, click the edit button. $\endgroup$ – DanielV Feb 25 '15 at 1:53

$$(n+1)!=(n+1) n! < (n+1) n^n<(n+1) (n+1)^n=(n+1)^{n+1} $$

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  • $\begingroup$ Oh cause you know (n)$^n < (n+1)^n$? for integers > 1? $\endgroup$ – committedandroider Feb 25 '15 at 1:54

You have $$k^{k+1} + k^k = k^k(k+1) $$

Can you use this fact?

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