Here is what I think might work:
Step 1: Base case. You've done this.
Step 2. Assume $4k<2^k$ for arbitrary k contained in the naturals. You can do this because your base case proves the assumption true for at least one k. That is, you've proved this true for k=5.
Step 3. Prove this holds for $n=k+1$.
$$4k<2^{k}$$ $$8k<2^k*2$$ $$8k<2^{k+1}$$
It is obvious that: $4k+1<8k<2^{K+1}$ for $K>4$
And therefore, by induction, that $4n<2^n$ for all n belonging to the naturals where n>4.
That is my attempt. Hope it works.