simple proof help (I think I'm really close)

Suppose that $\displaystyle h_0, h_1, h_2, h_3, \dots$ is a sequence defined as follows: $\displaystyle h_0 = 1, h_1 = 2, h_2 = 3, h_k = h_{k-1}+h_{k-2}+h_{k-3}$ for all ints $\displaystyle k \geq 3$. Prove that $\displaystyle h_n\leq 3^n $ for all ints $\displaystyle n \geq 0$

What I have so far is:

base case n=0, then $\displaystyle h_0=1$ and $\displaystyle 1 \leq 3$.

inductive step:

assume true for the base case, then $\displaystyle h_{n+1} = h_{(n+1)-1} + h_{(n+1)-2} + h_{(n+1)-3} = h_{n} + h_{n-1} + h_{n-2}$

since the base case is $\displaystyle \leq 3^n$, we really just want to show that our inductive case is $\displaystyle \leq 3^{n+1} = 3 \times 3^n$

but how can I show that $\displaystyle h_{n+1} \leq 3^{n+1}$?

any help with this would be greatly appreciated!