<s>[INST] 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 $n = p + 1$:
$$2^{3(k+1)} - 3{k + 1} = 2 * 2^p - 3$$
$$=2(5n + 3) - 3=10n + 6 - 3 = 10n+3$$
 [/INST] By I.H. we have $2^{3n}-3^{n}=5k$ for some integer $k$ and thus $2^{3n}=3^{n}+5k$.
Now we have:
$$2^{3(n+1)}-3^{n+1} = 8\cdot 2^{3n}-3^{n+1}= 8(3^{n}+5k)-3^{n+1} = 3^n(8-3)+40k = 5\cdot (3^n+8k)$$
and we are done.
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