Prove that n! yper extge 2ⁿ-1 for all n yper extge 1.

For $n=1$, we check if $P(1)$ is true.
Left Hand Side (LHS): $1! = 1$.
Right Hand Side (RHS): $2^{1-1} = 2^0 = 1$.
Since $1 \ge 1$, the statement $P(1)$ is true.

Assume that $P(k)$ is true for some positive integer $k \ge 1$.
This means we assume $k! \ge 2^{k-1}$.

We need to prove that $P(k+1)$ is true, i.e., $(k+1)! \ge 2^{(k+1)-1}$, which simplifies to $(k+1)! \ge 2^k$.
Consider the Left Hand Side of $P(k+1)$:

By the definition of factorial, $(k+1)! = (k+1) \cdot k!$.
From our inductive hypothesis, we know that $k! \ge 2^{k-1}$.
Substitute this into the expression:

Now, we need to show that $(k+1) \cdot 2^{k-1} \ge 2^k$.
We know that $2^k = 2 \cdot 2^{k-1}$.
So, we need to show that $(k+1) \cdot 2^{k-1} \ge 2 \cdot 2^{k-1}$.
Since $2^{k-1}$ is a positive value for $k \ge 1$, we can divide both sides of the inequality by $2^{k-1}$ without changing the direction of the inequality:

Since $k$ is a positive integer and $k \ge 1$, it follows that $k+1 \ge 1+1$, so $k+1 \ge 2$.
This inequality is true for all $k \ge 1$.
Therefore, we have shown that $(k+1)! \ge (k+1) \cdot 2^{k-1} \ge 2 \cdot 2^{k-1} = 2^k$.
Thus, $(k+1)! \ge 2^k$.
This proves that $P(k+1)$ is true.

By the principle of mathematical induction, the statement $n! \ge 2^{n-1}$ is true for all positive integers $n \ge 1$.