Prove $ \ \forall n \ge 100, \ n^{2} \le 1.1^{n}$ using induction. Question:
Prove $ \ \forall n \ge 100, \  n^{2} \le 1.1^{n}$ using induction. 
My attempt:
Base case is trivial.
Suppose $ \ n \ge 100$ and $ \ n^{2} \le 1.1^{n}$.
Then,
$(n+1)^{2} = n^{2} + 2n + 1 \le 1.1^{n} + 2n + 1$, by I.H.
I am stuck here. Any help would be appreciated. 
 A: Start from the RHS, use the hypothesis of induction and $n\ge 100$
$$1.1^{n+1}=(1.1)^n\cdot 1.1\ge 1.1n^2=n^2+0.1n^2\ge n^2+10n \ge n^2+2n+1=(n+1)^2$$
A: We note that for any $n \geq 100$, if we can prove that $\frac{(n+1)^2}{n^2}\leq 1.1$, then the claim should follow immediately.
$$
\frac{(n+1)^2}{n^2} = \left(\frac{n+1}{n}\right)^2 \leq \left(\frac{101}{100}\right)^2 = \frac{10201}{10000} < \frac{11000}{10000} = 1.1
$$
Note that the inequality $\frac{n+1}{n} \leq \frac{101}{100}$ can be seen by rewriting the LHS as $1 + \frac{1}{n}$ and the RHS as $1 + \frac{1}{100}$, and using the fact that $n \geq 100$.
A: Note that $n>25$ so $26n>25(n+1)$. Also note that $1.1> (\frac{26}{25} )^2$
\begin{eqnarray*}
(1.1)^{n+1}>1.1n^2>\left(\frac{26n}{25}\right)^2>(n+1)^2
\end{eqnarray*}
A: $$(n+1)^2=n^2\Bigl(1+\frac1n\Bigr)^2\le 1.1^n\Bigl(1+\frac2n+\frac1{n^2}\Bigr),$$
so all you have to prove is 
$$\frac 2n+\frac1{n^2}\le 0.1\iff n^2-20n-10\ge 0$$
As the positive root is  $10+\sqrt{110} <10+11$, this happens  for all $n\ge 21$,  and the inequality is established if we prove it's true for $n=100$. Actually, it's satisfied from $n=96$.
A: Other answers have provided the inductive step, so I will limit myself to proving the base case, $1.1^{100}\gt100^2$, in a way that can be checked by eye (i.e., without resorting to a calculator).
Since $100^2=10^4$, it suffices to show $1.1^{25}\gt10$.  We have
$$1.1^5=\left(1+{1\over10}\right)^5\gt1+5\cdot{1\over10}+10\cdot{1\over10^2}=1.6={8\over5}$$
and
$$\begin{align}
\left(8\over5\right)^5\gt10
&\iff2^{15}\gt5^5\cdot2\cdot5\\
&\iff2^{14}\gt5^6\\
&\iff2^7\gt5^3\\
&\iff128\gt125
\end{align}$$
Thus
$$1.1^{25}=(1.1^5)^5\gt\left(8\over5\right)^5\gt10$$
Remark:  As Bernard observes, the inequality $1.1^n\gt n^2$ actually starts at $n=96$.  It'd be nice to show this (and $1.1^{95}\lt95^2$) in a similar eyeball-checkable fashion, but I don't see any easy way to do so.
