Prove using Mathematical Induction that $2^{3n}-3^n$ is divisible by $5$ for all $n≥1$. I did most of it but I stuck here I attached my working 
tell me if I did correct or not thanks
My working:

EDITED: I wrote the notes as TEX
Prove using induction that $2^{3n} - 3^n \mod{5} = 0$.
Statement is true for $n = 1$: 
$$2^{3 * 1} - 3^1 = 2^3 - 3 = 8 - 3 = 5$$
$$5 \mod{5} = 0$$
Now for $n = p$ and $n = p + 1$:
$$2^{3(k+1)} - 3{k + 1} = 2 * 2^p - 3$$
$$=2(5n + 3) - 3=10n + 6 - 3 = 10n+3$$

We first show for the $n=1$ case:
$2^3-3 = 8-3 = 5$
which is clearly divisible by 5. Next is the inductive step where we assume 
$2^{3n} - 3^n=5m $ for some integer $m$.
We can rewrite this as 
$2^{3n}=5m+3^n$.
Now we want to prove that $2^{3(n+1)}-3^{n+1}$ is divisible by 5. We first can simplify this a bit:
$2^{3(n+1)}-3^{n+1} = 2^{3n+3}-3^{n+1}=2^3\cdot2^{3n}-3\cdot 3^n$.
We can now make the substitution from the previous statement to write this as
$2^3(5m+3^n)-3\cdot 3^n=5(8m)+8\cdot 3^n-3\cdot 3^n = 5(8m)+5\cdot3^n=5(8m+3^n)$
which is clearly divisible by 5 and proves the statement.
