I need to show that n^3 is less than or equal to n! Here is my attempt (or skip

Question

I need to show that n^3 is less than or equal to n!

Here is my attempt (or skip and show me another way)

Observe that n = 6 is sufficiently large and take n greater or equal to 6. Now assume to the contrary that n^3 > n! Then there is a smallest positive integer by WOP call this integer m. Therefore, m^3 > m! While n^3 is less than or equal to n! for every integer with 1 less than or equal to n < m. Since true for n = 2-6 it follow thats m is greater than or equal to 6. Hence we can write m = k+5 where 1 is less than or equal to k < m. Consequently k^3 > k! Now, m^3 > m! = (k+5)^3 > (k+5)!

This is where I am stuck. I could certainly expand everything out, and I suppose the right hand side would really be bigger and thus produce a contradiction.

Explanation / Answer

You can start with n=6 using induction. Remember we are trying to prove this via induction not contradiction. For n=6 6^3 < 6! 6*6*6 < 6*5*4*3*2*1 216 < 720 Now we assume that this holds true for n n^3

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