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How can you prove that for $n\geq1$, you have $n^n\geq e^{n-1}$? (no induction)

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Have you tried differentiation? If that doesn't work, try series expansion of e. –  The Chaz 2.0 Mar 26 '11 at 13:24
You can write $n^n= e^z$ for some $z \in \mathbb{R}$ and then compare exponents. –  Sebastian Mar 26 '11 at 13:30
what's wrong with induction? –  Andrea Mori Mar 26 '11 at 17:50

3 Answers 3

For $n\ge 3$ you clearly have $n^n \ge e^n \ge e^{n-1}$. (Since $e\le 3\le n$.) So it only remains to check the inequality for $n=2$.

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Well you can take logarithms, say

$$n^n \geq e^{n-1} \iff n\log{n} \geq n - 1 \iff n(\log{n} - 1) + 1 \geq 0$$

and the last inequality is true for any $n \geq 1$.

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Problem given: is $ n^n \ge e^{n-1} $ where $ n\ge 1 $ ?

Because $n \ge 1$ set $n=e^m $ so $m \ge 0$

Such substitution is often useful and you should always try whether this "simplifies by generalization". In this sense I would even write $n=e^{e^m}$ so that the parameter $m$ in the rewritten problem can vary over the whole real axis and still always satisfies the required $n$ - but this is not needed here.

$ \qquad \begin{array} {} e^{m e^m} &\ge& e ^{e^m-1} \\ m e^m &\ge& e^m - 1 \end{array} $

and then
$ \qquad m + m^2 + {m^3 \over2!} + \ldots \ge m+{m^2 \over 2!}+{m^3 \over 3!} + \ldots$

proves the assertion in the given problem by termwise comparision.

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