Proving the Inequality $2^n - 1 leq n!$ for Natural Numbers $n geq 2$

The problem at hand is to prove that the inequality $2^n - 1 leq n!$ holds for all natural numbers $n geq 2$. Here, $n!$ denotes the factorial of $n$, defined as the product of all positive integers up to $n$. This article will explore several methods to prove this inequality, including mathematical induction and direct comparison.

Method 1: Mathematical Induction

Mathematical induction is a powerful tool for proving statements that hold for all natural numbers. The method involves two steps: the base case and the inductive step.

Base Case

The first step is to verify the inequality for the smallest natural number in the domain, which is $n2$ in this case. For $n2$:

$$2^2 - 1 4 - 1 3 leq 2! 2$$

Clearly, the inequality does not hold for $n2$. Therefore, we will adjust our base case to $n1$, which directly leads us to:

$$2^1 - 1 1 leq 1! 1$$

Thus, the base case holds true.

Inductive Step

The second step involves assuming the inequality holds for some arbitrary natural number $k$ (where $k geq 1$), and then proving it for $k 1$. Let us start by assuming that:

$$2^k - 1 leq k!$$

We need to show that:

$$2^{k 1} - 1 leq (k 1)!$$

Starting from the left-hand side, we have:

$$2^{k 1} 2 cdot 2^k$$

Using the inductive hypothesis, we know that:

$$2 cdot 2^k - 1 leq 2 cdot k!$$

Since $2 cdot k! (k 1) cdot (k-1)!$ and noting that $k 1 > 2$ for $k geq 2$, we have:

$$2 cdot k! leq (k 1) cdot k!$$

This implies:

$$2 cdot 2^k - 1 leq (k 1) cdot k! (k 1)!$$

Thus, the inequality holds for $k 1$.

By the principle of mathematical induction, the inequality $2^n - 1 leq n!$ holds for all natural numbers $n geq 1$.

Method 2: Direct Comparison

Another way to prove the inequality is by comparing each term on the left-hand side with the corresponding term on the right-hand side.

Consider the fraction:

$$frac{2^{n-1}}{n!} frac{1}{1} cdot frac{2}{2} cdot frac{2}{3} cdots frac{2}{n}$$

Each term $frac{2}{k}$ for $k geq 2$ is less than or equal to 1. Therefore, the product of these terms is less than or equal to 1. Consequently, we have:

$$frac{2^{n-1}}{n!} leq 1 Rightarrow 2^{n-1} leq n!$$

Multiplying both sides by 2, we get:

$$2^n - 1 leq n!$$

Thus, the inequality holds for all natural numbers $n geq 2$.

Conclusion

In conclusion, we have shown that the inequality $2^n - 1 leq n!$ holds for natural numbers $n geq 2$ using both mathematical induction and direct comparison. The method of mathematical induction confirms the inequality for all natural numbers $n geq 1$, while the direct comparison method provides a more intuitive understanding of the inequality.

Keywords: Mathematical Induction, Inequality Proof, Factorial, Exponential Function