# Factorial inequation

##### New member
Hello,i hope everybody on this forum is fine, I'm actually in grade 12 and I need
Help for a inequation with factorial :
(K^(n+1))/K(n!) <= (K^K) / K!
Thank you
Ps sorry for my English it is not my mother langage...

#### Dr.Peterson

##### Elite Member
What is the goal? Are you trying to solve for K in terms of n, or for n in terms of K, or to prove it is always true, or what? I assume n and K are both natural numbers (positive integers).

Also, please tell us your context (what subject you are studying if this is for a class, and what you have learned that might be useful), and show us your thinking, so we can tell what help you need.

##### New member
What is the goal? Are you trying to solve for K in terms of n, or for n in terms of K, or to prove it is always true, or what? I assume n and K are both natural numbers (positive integers).

Also, please tell us your context (what subject you are studying if this is for a class, and what you have learned that might be useful), and show us your thinking, so we can tell what help you need.
You are right I'm sorry, I want to prove that this is always true, I forgot to add, like you say, that N is a natural number and K also and K > n
I'm a student in Japan and I am doing Exercice in recurrence thinking (I don't know if my translation is true) but it is basically when you suppose a statement for a n number and you prove it with n+1 usually used in sequence...
Thank you

##### New member
What is the goal? Are you trying to solve for K in terms of n, or for n in terms of K, or to prove it is always true, or what? I assume n and K are both natural numbers (positive integers).

Also, please tell us your context (what subject you are studying if this is for a class, and what you have learned that might be useful), and show us your thinking, so we can tell what help you need.
Edit : I changed my previous reply because I did a mistake it is k>n not k>0!
Thank you

#### ksdhart2

##### Senior Member
To me this almost feels like a bit of a "trick" question, in that it seems really complex but it's actually not. You don't even need to do a proof by induction (the concept you were describing in post #3), as you can directly demonstrate it's true for any pair $$(K, n)$$ that satisfies $$K \gt n$$. Work on the left-hand side and divide top and bottom by $$K$$:

$$\displaystyle \frac{K^{n+1}}{K \times n!} = \frac{\left( \frac{K^{n+1}}{K} \right)}{\left( \frac{K \times n!}{K} \right)}$$

For this step to be valid, you do have to assume that $$K \neq 0$$, but that's already known because $$K$$ and $$n$$ are natural numbers and $$K \gt n$$ ensures $$K \geq 1$$ anyway. Once you've done this step, the rest of the proof should be clear.

 To me this almost feels like a bit of a "trick" question, in that it seems really complex but it's actually not. You don't even need to do a proof by induction (the concept you were describing in post #3), as you can directly demonstrate it's true for any pair $$(K, n)$$ that satisfies $$K \gt n$$. Work on the left-hand side and divide top and bottom by $$K$$: $$\displaystyle \frac{K^{n+1}}{K \times n!} = \frac{\left( \frac{K^{n+1}}{K} \right)}{\left( \frac{K \times n!}{K} \right)}$$ For this step to be valid, you do have to assume that $$K \neq 0$$, but that's already known because $$K$$ and $$n$$ are natural numbers and $$K \gt n$$ ensures $$K \geq 1$$ anyway. Once you've done this step, the rest of the proof should be clear. Click to expand... Thank you very much, I have prove it thanks to your help. Have a nice day