Explanation
We have,
$$P\left( n \right)$$ be the statement $${{2}^{n}}<n!$$
Where $$n$$ is a natural number
Then,
Put $$n=1,2,3....$$
So,
$$P\left( 1 \right)\,$$ be the statement of $${{2}^{1}}<1!=2<1\,\,\,\left( \text{It}\,\text{is}\,\text{wrong} \right)$$
$$P\left( 2 \right)$$ be the statement of $${{2}^{2}}<2!=4<2\,\,\left( \text{It}\,\text{is}\,\text{wrong} \right)$$
$$P\left( 3 \right)$$ be the statement of $${{2}^{3}}<3!=8<6\,\,\left( \text{It}\,\text{is}\,\text{wrong} \right)$$
But,
$$P\left( 4 \right)$$ be the statement of $${{2}^{4}}<4!=16<24\,\,\left( \text{It}\,\text{is}\,\text{right} \right)$$
Similarly,
$$P\left( 5 \right),\,P\left( 6 \right)\,.......$$ So, all number $$n$$ is a natural number.
Therefore, it is true for all $$n>3$$.
Hence, this is the answer.
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