Using Induction to prove that $n^3log^4n=O(n^4)$ I need to prove the following asymptotic relation for the purpose of cacluating a recurrence relation:
$$n^3log^4n=O(n^4)$$
I tried and failed to do it with induction, which, if possible using basic Calculus 1 level math, I would like you to help me with.
I am not required to do it with induciton or anything, I just wondered if I can do it.

I was able to prove it though, in a differet manner:
$$f \in o(g) \Rightarrow f \in O(g)$$
$$lim_{n\to \infty} \frac{n^3log^4n}{n^4}=0$$
Using L'Hôpital's rule 4 times, which proves that:
$$n^3log^4n=o(n^4)$$
By the definitino of $o$.

Therefore it also follows that $$n^3log^4n=O(n^4)$$

For the basis of the induction let $n=1$ and with $c=1$:
$$f(1) = 1^3log^4(1) = 0 \leq 1=1^4 \cdot 1=c \cdot g(1)$$
From here I think that it would be enough to show that:
$$log^4(n) \leq n$$
Although I am not sure how to continue from here.
You showed that f (n) = o (g (n)). That's it. There is nothing else to prove. 

If you look at the definition of o (f (n)) and O (f (n)), they are almost identical except one says "for every eps > 0" and the other says "there is one c > 0". You can take every single eps of the little-o definition and use it as the c in the big-O definition.