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So I'm trying to prove, by induction, that $$ n^n \geq n!, \forall n\geq1$$

Base case:

$$ \text{For } n=1, 1^1 = 1 \geq 1 = 1!$$

Hypothesis:

$$ n^n \geq n!$$

Step:

$$ \text{Trying to prove: } n^{n+1} \geq (n+1)! $$

Now, somewhere around here I get some contradicting things. For example, if I start from the right side I get:

$$ (n+1)! = (n+1)\cdot n! \leq (n+1)\cdot n^n = n\cdot n^n + n^n = n^{n+1} + n^n$$

Based on this I would need $n^{n+1} + n^n$ to be less than or equal to $n^{n+1}$, which is certainly not true. Something similar happens when I go the other way.

Any ideas what I'm doing wrong here? Thanks.

Did
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Koy
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    "Trying to prove: $n^{n+1} \geq (n+1)!$" No: Trying to prove: $(n+1)^{n+1} \geq (n+1)!$ – Did Dec 14 '18 at 12:29
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    Well, now I feel stupid. Thanks. – Koy Dec 14 '18 at 12:30
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    Do you have to prove this with induction? I think you could just write out what the two terms mean, and you will see that the hypothesis is true. – Matti P. Dec 14 '18 at 12:30
  • Indeed induction is not at all the most direct route here. – Did Dec 14 '18 at 12:31
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    "Well, now I feel stupid." Who never does? "Thanks." You are welcome. – Did Dec 14 '18 at 12:31
  • You forget $n+1$ in the base... of your 1 step... – dmtri Dec 14 '18 at 12:33
  • None of the existing answers have pointed out that your getting contradiction does not imply a problem. In fact, $n^{n+1}$ IS greater than $(n+1)!$ for every $n ≥ 2$, which is immediately obvious from $n^3 > 1·2·(n+1)$. – user21820 Jan 26 '19 at 14:40

5 Answers5

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You should try to prove that $$(n+1)^{n+1} \ge (n+1)!$$

\begin{align} (n+1)^{n+1} &= (n+1) (n+1)^n\\ &\ge(n+1)n^n \end{align}

Now use induction hypothesis.

Siong Thye Goh
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You are trying to prove :

$$(n+1)^{n+1} \geq (n+1)!$$

Not :

$$n^{n+1} \geq (n+1)!$$

Thinking
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You need to show that $(n+1)^{n+1} \geq (n+1)!$, not that $n^{n+1} \geq (n+1)!$.

Here are some similar questions asked before: you can check your work against any of them if you'd like.

For future reference, Approach0 is an excellent resource to search for similar questions (much better than the SE functionality itself). All the best.

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$(n+1)! = (n+1)\cdot n! \leq (n+1)\cdot n^n$ by induction hypothesis, and $(n+1)\cdot n^n\leq (n+1)\cdot (n+1)^n = (n+1)^{n+1}$. Done.

Wuestenfux
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As an alternative: $$ n! = \underbrace{{n(n-1)(n-2)\cdots 3\cdot 2\cdot 1}}_{n\ \text{times}} \\ n^n = \underbrace{n \cdot n\cdot n \dots n}_{n\ \text{times}} $$

Note that: $$ {n! \over n^n} = \frac{n(n-1)(n-2)\cdots 3\cdot 2\cdot 1}{n \cdot n\cdot n \dots n} = \\ = \frac{n}{n} \cdot \frac{n-1}{n} \cdot \frac{n-2}{n} \cdots \frac{2}{n} \cdot \frac{1}{n} $$

Now note that for any $n \ge 2$: $$ \frac{n!}{n^n} < 1 $$

roman
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