Proving $n^n\geq e^{n-1}$ How can you prove that for $n\geq1$, you have $n^n\geq e^{n-1}$? (no induction)
 A: 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 $n \geq 1$.
A: 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 $n=2$.
A: 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.    
Then
$ \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.
