A simple proof by induction $P(n) = \frac{6^{2n} - 3^n}{11} \in \mathbb{N}$

You don't really need induction, or just use some simple production, as $$6^{2n}-3^n=3^n(12^n-1)$$

We have $P(n)=\frac{6^{2n}-3^n}{11} \in \mathbb{N}$.

You have proven that the base case is true.

Now, assume true for $n=k$.

$$6^{2k}-3^k=11p \tag{1}$$

Where $p \in \mathbb{N}$.

True for $n=k+1$.

$6^{2(k+1)}-3^{k+1}=6^{2k+2}-3^{k+1}=36 \cdot 6^{2k}-3 \cdot 3^k \tag{2}$

Now, here comes the trick.

We can substitute from $6^{2k}$ from equation $(1)$ into equation $(2)$.

$$36 \cdot (11p+3^k)-3 \cdot 3^k=11\cdot 36p+36\cdot 3^k-3 \cdot 3^k=11 \cdot 36p-33 \cdot 3^k \tag{3}$$

Since $11\cdot 36p$ is obviously divisible by $11$, and $33 \cdot 3^k$ can be divided by $11$ to give $3 \cdot 3^k$, the whole of equation $(3)$ is divisible by $11$.

Hence, we are done.

Because $\displaystyle \frac{6^{2n}-3^n}{11}$ is a natural number, we can write for some $k \in \mathbb N, 6^{2n} =3^n +11k $. Then our proof becomes $$\frac {6^{2n+2}-3^{n+1}}{11} = \frac {36 (3^n +11k )-3 (3^{n})}{11} =\frac {33 (3^n) +11 (36k)}{11} \in \mathbb N $$ And thus the proof is completed. Hope it helps.