Proof verification of $6^n \geq n3^n$ n is a natural number. Prove $6^n \geq n3^n$ holds for every natural number.
$n = 1:$
$$6 \geq 3 $$
$n \rightarrow n + 1:$
$$6 ^{n+1} = (3*2)^{n+1} = 3^{n+1}2^{n+1} \geq 3^{n+1}(n+1)2\geq 3^{n+1}(n+1) $$
Is my proof correct?
EDIT:
$$2^{n+1} \geq(n+1)2 \text{ because of Bernoulli inequality}$$

Suppose $6^n\ge n\cdot 3^n$ is true for $n$ and let's prove it for $(n+1)$.
$6^{n+1}=6\cdot 6^n\ge 6\left(n\cdot 3^n\right)=2\cdot 3\left(n\cdot 3^n\right)=2 n\cdot 3^{n+1}\ge (n+1)3^{n+1}$
proved.
