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 $2^{100}\gt 100^{10}$ ?

If $n\ge100$, then by binomial expansion
$$
\begin{aligned}
(n+1)^{10}&=n^{10}+10n^9+45n^8+120n^7+210n^6+252n^5+210n^4+120n^3+45n^2+10n+1\\
&\le n^{10}+10n^9+(45+120+210+252+210+120+45+10+1)n^8\\
&=n^{10}+10n^9+1013n^8\quad\text{(as $n^k<n^8$ when $k<8$)}\\
&\le n^{10}+10n^9+11n^9\quad\text{(as $1013<11n$)}\\
&=n^{10}+21n^9\\
&<n^{10}+100n^9\\
&\le 2n^{10}.
\end{aligned}
$$
A simple induction gives the rest. If $n^{10}<2^n$ for some $n\ge100$, then the above shows that
$$(n+1)^{10}<2n^{10}<2\cdot2^n=2^{n+1}$$
completing the inductive step. The base case $2^{100}>100^{10}$ is routine. For example, starting from $2^{10}=1024>10^3$ we immediately get that $2^{100}>10^{30}>100^{10}$.
