$3^{3n+1} < 2^{5n+6} $ for all non-negative integers $n$. Is my induction solution correct? Show using mathematical induction that $3^{3n+1} < 2^{5n+6} $ for all non-negative integers $n$. I'm not sure whether what I did at the last is valid? 
Basis step: 
for all non-negative integers
$$P(n) = 3^{3n+1} < 2^{5n+6} $$ 
$$P(0) = 3^{3(0) + 1} = 3 < 64 = 2^{5(0) + 6}$$
$$P(0) = T$$
Inductive Step:
Assume: $3^{3k+1} < 2^{5k+6}$
Show: $3^{3(k+1)+1} < 2^{5(k+1)+6}$
$$ 3^{3(k+1)+1} = 3^{3k+4} = 3^3 \cdot 3^{3k+1}$$
By inductive hypothesis~
$$3^3 \cdot 3^{3k+1} < 3^3 \cdot 2^{5n+6} $$
This is the part where I'm not sure if you can do this in induction but it seems logically correct.
$$3^3 \cdot 2^{5n+6} = 27 \cdot 2^{5n+6}$$
$$2^{5(k+1)+6} = 2^{5k+5+6}= 2^5 \cdot 2^{5n+6} = 32 \cdot 2^{5n+6}$$
I'm not sure whether it should be $\le$ or $<$ but I used '$<$' for $3^3 \cdot 2^{5n+6}<2^{5(k+1)+6} $
Therefore: 
$$3^{3(k+1)+1} <  3^3 \cdot 2^{5n+6}<2^{5(k+1)+6} $$

After your first invokation of the inductive hypothesis you reach
$3^3 \cdot 3^{3k+1} < 3^3 \cdot 2^{5k+6}$
So why not just direction show that $3^3 < 2^5$, completing the proof?
$3^3 \cdot 2^{5k+6} < 2^5 \cdot 2^{5k+6} = 2^{5(k+1)+6}$
