Math Induction Proof: $(1+\frac1n)^n < n$ So I have to prove: 

For each natural number greater than or equal to 3,
  $$(1+\frac1n)^n<n$$

My work:
Basis step: $n=3$
$$\left(1+\frac13\right)^3<3$$
$$\left(\frac43\right)^3<3$$
$$\left(\frac{64}{27}\right)<3$$
which is true.
Now the inductive step, assume $P(k)=\left(1+\frac1k\right)^k<k$ to be true and prove $P(k+1)=\left(1+\frac1{k+1}\right)^{k+1}<k+1$. 
This is where I am stuck because usually you add or multiply by $k+1$ or some similar term.

Two approaches

Bernoulli-like Inductive Step
$$
\begin{align}
\frac{\left(1+\frac1{n+1}\right)^{n+1}}{\left(1+\frac1n\right)^n}
&=\overbrace{\left(\frac{n(n+2)}{(n+1)^2}\right)^{n+1}}^{\lt1}\frac{n+1}n\\
&\lt\frac{n+1}n
\end{align}
$$
This shows that if $\left(1+\frac1n\right)^n\lt n$, then $\left(1+\frac1{n+1}\right)^{n+1}\lt n+1$

Non-Inductive Bernoulli Approach
Bernoulli's Inequality says that for $n\ge2$,
$$
\begin{align}
\left(1-\frac1{n+1}\right)^n
&=\left(\left(1-\frac1{n+1}\right)^{n/2}\right)^2\\
&\ge\left(1-\frac{n/2}{n+1}\right)^2\\[3pt]
&=\left(\frac{n/2+1}{n+1}\right)^2
\end{align}
$$
which is the reciprocal of
$$
\begin{align}
\left(1+\frac1n\right)^n
&\le\left(\frac{n+1}{n/2+1}\right)^2\\[6pt]
&\lt2^2
\end{align}
$$
Thus, $\left(1+\frac1n\right)^n\lt n$ if $n\ge4$.
We just need to verify the inequality for $n=3$:
$$
\left(1+\frac13\right)^3=\frac{64}{27}\lt3
$$
