I really need help with this proof. Here’s what I have tried: n^6 <= (7^n)c is true for n=1,c=1
So I want to prove (n+1)^6 <= (7^n+1)
I’ve tried expanding n+1^6 term and manipulating into the form 7^n(7) but I’m stuck and see no other similar questions online.
Statement $$f(n) = O(g(n))$$ means $$\sup_{n \ge 1} \left|\frac{f(n)}{g(n)}\right| < \infty $$
Since $$\lim_{n\to\infty}\frac{n^6}{7^n}=0$$ it automatically implies that $$\sup_{n \ge 1} \frac{n^6}{7^n} < \infty $$
therefore $n^6=O(7^n)$
For $n\ge 10$ we have $n-5\ge {n\over 2}.$ Thus $$7^n> 2^n>{n\choose 6}>{n(n-5)^5\over 6!}\ge {n^6\over 2^56!} $$ Hence $${n^6\over 7^n}<2^56!,\quad n\ge 10$$ You can as well show that $${n^6\over 7^n}\ge {(n+1)^6\over 7^{n+1}},\quad n\ge 6$$ Hence $${n^6\over 7^n}\le {6^6\over 7^6}<1,\quad n\ge 6$$
For your induction you can proceed like this for $n\ge 6$:
$$(n+1)^6=n^6\Big(1+\frac 1n\Big)^6=n^6\Big(1+\underbrace{\frac 6n}_{\le 1}+\underbrace{\frac{15}{n^2}}_{\le 1}+\underbrace{\frac{20}{n^3}}_{\le 1}+\underbrace{\frac{15}{n^4}}_{\le 1}+\underbrace{\frac{6}{n^5}}_{\le 1}+\underbrace{\frac 1{n^6}}_{\le 1}\Big)\le 7n^6$$
