How can I prove that $n^{55} = O(2^n)$ using the following Big O definition: $$ f(g(n)) = O(g(n)) \\ \Leftrightarrow $$ there are positive constants $c$ and $n_0$, such that $|f(g(n))| \le c \cdot g(n)$, for all $n \ge n_0$.
I know that $n^c$ will always grow slower than $2^n.$
So we start with: $$ 10n^{55} \le c \cdot 2^n. $$
Then, I used logarithms but without success :(
I also thought of some how turn the exponential function into a polynomial one, or vice versa, but as a computer science student, I have no idea how to do that.
Hint:
$$\left(1+\frac1{79}\right)^{55}<2$$
so that for all $n\ge79$,
$$\left(\frac{n+1}n\right)^{55}<2$$
and by the telescoping product, $n^{55}$ grows slower than $2^n$.