Prove that $2^n + 5^n + 56$ is divisible by $9$, where $n$ is an odd integer 
Prove that $9 \mid2^n + 5^n + 56$ where n is odd

I have proved this by using division into cases based on the value of $n\bmod3$ but it seems a little bit clumsy to me and I wonder if there are other ways to prove it, probably by using modular arithmetic or induction? Below is my proof:
$\text{Case 1, }n\bmod3=0,\text{then $n=3k$ for some odd integer k:}$ $$\begin{align}
2^n+5^n+56 & =2^{3k}+5^{3k}+56
\\ & = 8^k+125^k+56
\\ & \equiv (-1)^k+(-1)^k+2\quad&\left(\bmod9\right)
\\ & \equiv 0\quad&\left(\bmod9\right)
\end{align}$$
$\text{Case 2, }n\bmod3=1,\text{then $n=3k+1$ for some even integer k:}$ $$\begin{align}
2^n+5^n+56 & =2^{3k+1}+5^{3k+1}+56
\\ & = 2\cdot8^k+5\cdot125^k+56
\\ & \equiv 2\cdot(-1)^k+5\cdot(-1)^k+2\quad&\left(\bmod9\right)
\\ & \equiv 9\equiv0\quad&\left(\bmod9\right)
\end{align}$$
$\text{Case 3, }n\bmod3=2,\text{then $n=3k+2$ for some odd integer k:}$ $$\begin{align}
2^n+5^n+56 & =2^{3k+2}+5^{3k+2}+56
\\ & = 4\cdot8^k+25\cdot125^k+56
\\ & \equiv 4\cdot(-1)^k+25\cdot(-1)^k+2\quad&\left(\bmod9\right)
\\ & \equiv -27\equiv0\quad&\left(\bmod9\right)
\end{align}$$

Mathematical induction approach
We know that $n$ is an odd integer, i.e. we can use $n= 2k - 1, k \in \mathbb{N}$. Hence, we can reformulate a starting statement:
$$9 \mid 2^{2k-1} + 5^{2k-1} + 56, \quad k \in \mathbb{N}.$$
$\text{1.}$ Basis of Induction:
For $\ k=1 \ $ we have:
$$\ 2^1 + 5^1 + 56 = 63 = 9 \cdot 7, \ $$
i.e. the statement is true for $\ k=1$.
$\text{2.}$ Induction Hypothesis:
Suppose the statement holds for some $\ k>1, \ $i.e. that it holds:
$$\ 2^{2k-1} + 5^{2k-1} + 56 = 9 \cdot m, \ m \in \mathbb{N}.$$
$\text{3.}$ The proof:
Let's check if the statement holds for $\ k+1. \ $ We have:
$2^{2(k+1)-1} + 5^{2(k+1)-1}+56 = 2^{2k+2-1} + 5^{2k+2-1}+56=4\cdot 2^{2k-1} + 25 \cdot 5^{2k-1} + 56 =
\\ 4\cdot (2^{2k-1}+5^{2k-1}+56) + 21\cdot 5^{2k-1} - 168 = 4\cdot 9 \cdot m + 21 \cdot (5^{2k-1} - 8).$
The first summand on the right side (i.e. $\ 4\cdot 9 \cdot m$) is obviously divisible by $\ 9, \ $ and because we know $\ 21 = 3 \cdot 7, \ $ it remains to prove that it holds: $\ 3 \mid 5^{2k-1} - 8, \ k \in \mathbb{N}.$ For that purpose we will use a mathematical induction too!
