So, here I'm asking for a check..

For each natural number $n$ with $n\geq3$, $\left(1+\frac1{n}\right)^n<n$.

Proof:

Basis $n=3$: true.

Assume for some $k$ with $k\geq3$, that $\left(1+\frac1{k}\right)^k<k$. Multiplying by $\left(1+\frac1{k}\right)$ yields $\left(1+\frac1{k}\right)^{k+1}<k+1$. But since $k+1>k\implies{1/(k+1)<1/k}\implies{}\left[1+\frac1{k+1}\right]^{k+1}<\left(1+\frac1{k}\right)^{k+1}<k+1$ as desired.

the set $T$ of natural numbers which make the above statement true is such that $3\in{}T$, and if $k\in{}T,~k\geq3$, then $k+1\in{}T$.

Therefore, by the principle of induction, for every natural number $n\geq3$, $\left(1+\frac1{n}\right)^n<n$.

The part I worry about is when I show that k<k+1... Is this ok?