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This is my work so far but I am currently stuck and is a tad bit clueless on the steps to take Please try to prove using induction.

Question : Let $P(n)$ be the statement $n^2 \leq n!$ where $n$ is a non-negative integer. Explain the steps to be taken if you were to prove that $P(n)$ is true for any $n \geq 4$.

Attempt :

$P(n) = n^2 \leq n!$

$n \geq 4$

Base claim : If $n$ is $4$ ...

$4^2 \leq 4!$

$16\leq 4 \times 3 \times 2 \times 1$

$16 \leq 24$

Assume that

$P(k)$ is true, prove that $P(k+1)$ is true.

$P(k) : k^2 \leq k!$

Sarvesh Ravichandran Iyer
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PrivatePython
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  • Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. – Community Sep 30 '21 at 10:09
  • Try checking some more examples. Do you believe the claim? Why? – Karl Sep 30 '21 at 10:12
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    How is $(k+1)!$ related to $k!$? How is $(k+1)^2$ related to $k^2$? – Karl Sep 30 '21 at 10:13
  • They are related as they are technically one step after K^2 and K! but i do not know how to fit K+1 into it to proof it is true – PrivatePython Sep 30 '21 at 10:32
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    Per the comment of @Karl, suppose that you have two functions, $f(n)$ and $g(n)$. Further suppose that you know that $f(4) < g(4)$ and that you wish to prove that $f(n) < g(n)$, for all $n \in \Bbb{Z_{\geq 4}}.$ One approach is to contrast the $\displaystyle \frac{f(n+1)}{f(n)}$ against $\displaystyle \frac{g(n+1)}{g(n)}$. Do you understand why this comparison could be relevant? – user2661923 Sep 30 '21 at 10:34
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    Re previous comment, suppose that $f(N) < g(N)$ for some positive integer $N \geq 4$. Further suppose that $\displaystyle A = \frac{f(N+1)}{f(N)} < \frac{g(N+1)}{g(n)} = B.$ Then, $f(N+1) = A \times f(N) < B \times g(N) = g(N+1).$ – user2661923 Sep 30 '21 at 10:37
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    Note that if $f(N) = N^2$, then $\displaystyle \frac{f(N+1)}{f(N)} = \frac{(N+1)^2}{N^2}.$ As $N$ increases, the fraction $\displaystyle \frac{(N+1)^2}{N^2}$ is decreasing. Further $\displaystyle \frac{(4+1)^2}{4^2} < 2 < (4+1).$ – user2661923 Sep 30 '21 at 10:41
  • Note : rewritten the MathJax , I left out something in the right side column that looked to be a rewrite of $P(k)$, so I hope I haven't been too creative. Can be rolled back. Someone can take the hint above and create an answer. – Sarvesh Ravichandran Iyer Sep 30 '21 at 10:57
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    It looks like Inequalities - proof by induction that $n^2 \leq n!$ for $n\geq 4$ contains the answer to your question. – Sarvesh Ravichandran Iyer Sep 30 '21 at 11:01
  • The original question doesn't specify proof by induction, so consider the behaviour of $x(x-1)(x-2) - x^2$ – Arthur Vause Sep 30 '21 at 12:21

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