Prove by Mathematical induction for n > 6, n! > 3^n where n is a natural number

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Explanation / Answer

1)take n=7 we have 7! = 5040 3^7= 2187 2)Now suppose for k such that k>6 that n! > 3^n is valid 3)for k+1th term we have k+ 1! =k!*(k+1) and 3^(k+1) =3^(k)*3 now K! > 3^K => (k+1)!>3^(k+1) hence proved Now understand how this prove works we prove by calculation that the given statement is valid for the smallest number on which the condition is applied..(n=7 in this case) . Then we suppose that the given statement is true for a k in given condition (note that we already have one K=7). Now we extend the definition to k+1 using the our assumption for k. so in our case the statement was true for 7 we assume k=7 this gives us statement is true for 8. Now take k=8 again we get statement is true for 9 ......

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