<s>[INST] How would you prove inequality $2^n \gt n^{10}$ using induction For the base case I can put a number such as $100$ for $n$ so $2^{100}\gt 100^{10}$.
Ok so now the induction hyp:
$2^{n+1} > (n+1)^{10}$ for $n \gt 101.$
where do I go from here? Also do I have to prove $2^{100}\gt 100^{10}$ ?
 [/INST] We have $2^{n+1} = 2^n \cdot 2 \geq n^{10} \cdot 2$. From that we get 
$$(1+n)^{10} =\sum_{k=0}^{10}\binom{10}{k}n^k =n^{10} + \sum_{k=0}^{9}\binom{10}{k}n^k $$ and substracting $n^{10}$ from both sides left is to show that
$$\sum_{k=0}^{9}\binom{10}{k}n^k \leq n^{10},$$ dividing by $n^{10}$ obtains 
$$\frac{10n^9}{n^{10}} + \frac{45n^8}{n^{10}} +\frac{120n^7}{n^{10}} +\frac{210n^6}{n^{10}}  +\frac{252n^5}{n^{10}}  +\frac{210n^4}{n^{10}}  +\frac{120n^3}{n^{10}}  +\frac{45n^2}{n^{10}}  +\frac{10n}{n^{10}}  +\frac{1}{n^{10}} \leq 1  $$
which is
$$(*) =\frac{10}{n}+\frac{45}{n^2}+\frac{120}{n^3}+\frac{210}{n^4}+\frac{252}{n^5}+\frac{210}{n^6}+\frac{120}{n^7}+\frac{45}{n^8}+\frac{10}{n^9}+\frac{1}{n^{10}} \leq 1 $$  and since $n \geq 100$ we get 
$$(*) \leq \frac{10}{100}+\frac{45}{100^2}+\frac{120}{100^3}+\frac{210}{100^4}+\frac{252}{100^5}+\frac{210}{100^6}+\frac{120}{100^7}+\frac{45}{100^8}+\frac{10}{100^9}+\frac{1}{100^{10}} \leq 1$$
which can be verfied by calculating it. This completes the proof.
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