Prove that $(1+\frac{1}{n})^k < 1+ \frac{k}{n}+\frac{k^2}{n^2}$ Prove that $(1+\frac{1}{n})^k < 1+ \frac{k}{n}+\frac{k^2}{n^2}, \forall$ n, k nonnegative integers, $k\le n$.
I know that has something to do with Bernoulli's inequality $(1+\alpha)^x\ge 1+\alpha x, \alpha \ge -1, n\ge1$.
If I reconsider Bernoulli's inequality with $\alpha=\frac{1}{n}$ and $x=k$ it follows that
$(1+\frac{1}{n})^k\ge 1+\frac{k}{n}$, but I don't know how to continue.
I also tried to prove it with induction where I consider
$p(n):(1+\frac{1}{n})^k < 1+ \frac{k}{n}+\frac{k^2}{n^2}$ to be true and prove that $ p(n+1):(1+\frac{1}{n+1})^k < 1+ \frac{k}{n+1}+\frac{k^2}{(n+1)^2}$ to be also true, but it didn't work.
Thank you!

Proof without induction/calculus:
It's easy to prove the inequality is true for $k=1,2$. If $k\ge 3$ we have
$$\left(1+\frac{1}{n}\right)^k - 1- \frac{k}{n}-\frac{k^2}{n^2} $$ $$= - \frac{k^2}{n^2} + \binom{k}{2} \frac{1}{n^2} + \sum_{i=3}^k \binom{k}{i} \frac{1}{n^i} = -\frac{k(k+1)}{2n^2}+\sum_{i=3}^k \binom{k}{i} \frac{1}{n^i} \tag1$$
Notice that
$$\frac{\binom{k}{i+1}/n^{i+1}}{\binom{k}{i}/n^i}=\frac{k-i}{n(i+1)} \le \frac{k-3}{4n}, \forall i \ge 3$$
Then $(1)$ is less then
$$-\frac{k(k+1)}{2n^2} + \frac{\binom{k}{3}/n^3}{1-\frac{k-3}{4n}}$$ and it suffices to prove
$$ \frac{\frac{k(k-1)(k-2)}{6n^3}}{1-\frac{k-3}{4n}} < \frac{k(k+1)}{2n^2}$$
Or equivalently
$$\frac{4(k-1)(k-2)}{4n-k+3} < 3(k+1) \iff 4(k-1)(k-2) < 3(k+1)(4n-k+3)\\
\stackrel{k\le n}{\Leftarrow} 4(k-1)(k-2) < 3(k+1)(4k-k+3) = 9(k+1)^2$$ which is trivially true.
