<s>[INST] How to resolve $n>(1+\frac{1}{n})^n$? I'm trying to prove that $\forall n\geq 3, n^{n+1}>(n+1)^n$. I came that this is true for $n>(1+\frac{1}{n})^n$. WolphramAlpha gives $n>2.293166...$ but I failed to compute it analytically.
 [/INST] Crostul's solution is probably the most straightforward way to go about it, but I'm going to do some induction because I probably have nothing better to do.
Base Case: When $n = 3$, $n > \left(1 + \frac{1}{n}\right)^n$ since $3 > \frac{64}{27}$.
Inductive Step: Suppose that $n > \left(1 + \frac{1}{n}\right)^n$. We wish to show that $n + 1 > \left(1 + \frac{1}{n+1}\right)^{n+1}$.
Notice that with $n \ge 3$, we have $1 + \frac{1}{n+1} < 1 + \frac{1}{n}$.
Thus, if $n > \left(1 + \frac{1}{n}\right)^n$, then clearly $n > \left(1 + \frac{1}{n+1}\right)^n$.
It follows that $n\left(1 + \frac{1}{n}\right) > \left(1 + \frac{1}{n+1}\right)^n\left(1 + \frac{1}{n}\right) \Longleftrightarrow n+1 > \left(1 + \frac{1}{n+1}\right)^n\left(1 + \frac{1}{n}\right)$
We see that since $1 + \frac{1}{n+1} < 1 + \frac{1}{n}$, we have $\left(1 + \frac{1}{n+1}\right)^n\left(1 + \frac{1}{n}\right) > \left(1 + \frac{1}{n+1}\right)^{n+1}$. 
Therefore, $n + 1 > \left(1 + \frac{1}{n+1}\right)^{n+1}$.

Since we have shown a base case for $n = 3$ and have also shown that $n > \left(1 + \frac{1}{n}\right)^n \Longrightarrow n + 1 > \left(1 + \frac{1}{n+1}\right)^{n+1}$, we have proven that for $n \ge 3$, $n > \left(1 + \frac{1}{n}\right)^n$
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