Let $P(n)$ be a statement and let $P(n) \implies P(n + 1)$ for all natural numbers $n$. Then $P(n)$ is true for:

If $P(n): 2^{n} < n!$,then the smallest positive integer $n$ for which $P(n)$ is true is:

Prove the following by using the principle of mathematical induction for all $n \in N$ where $n \geq 2$:
$\left(1-\frac{1}{2^{2}}\right)\left(1-\frac{1}{3^{2}}\right)\left(1-\frac{1}{4^{2}}\right) \ldots \left(1-\frac{1}{n^{2}}\right)=\frac{n+1}{2n}$

Prove the statement by the Principle of Mathematical Induction: $3^{2n} - 1$ is divisible by $8$ for all natural numbers $n$.

Prove the statement by the Principle of Mathematical Induction:
$2n < (n+2)!$ for all natural numbers $n$.

Prove the following by using the principle of mathematical induction for all $n \in N$ :
$\frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{3 \cdot 4 \cdot 5} + \ldots + \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}$