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I am rather new to mathematical induction. Specially inequalities, as seen here How to use mathematical induction with inequalities?. Thanks to that question, I’ve been able to solve some of the form $ 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} \leq \frac{n}{2} + 1 $.

Now, I was presented this, for $n \ge 4$:

$$2^n<n!$$

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I tried to do it with similar logic as the one suggested there. This is what I did:

Prove it for $n = 4$:

$$2^4 = 16$$

$$4! = 1\cdot2\cdot3\cdot4 = 24$$

$$16 < 24$$

Assume the following:

$$2^n<n!$$

We want to prove the following for $n+1$:

$$2^{n+1}<(n+1)!$$

This is how I proved it:

- So first we take $2^{n+1}$ which is equivalent to $2^n\cdot2$
- By our assumption, we know that $2^n\cdot2 < n!\cdot2$
- This is because I just multiplied by $2$ on both sides.

- Then we’ll be finished if we can show that $n! \cdot 2 < (n+1)!$
- Which is equivalent to saying $n!\cdot2<n!\cdot(n+1)$
- Since both sides have $n!$, I can cancel them out
- Now I have $2<(n+1)!$
- This is clearly true, since $n \ge 4$

Even though the procedure seems to be right, I wonder:

- In the last step, was it ok to conclude with $2<(n+1)!$? Was there not anything else I could have done to make the proof more “careful”?
- Is this whole procedure valid at all? I ask because, well, I don’t really know if it would be accepted in a test.
- Are there any points I could improve? Anything I could have missed? This is kind of the first time I try to do these.

- Prove that $ n < 2^{n}$ for all natural numbers $n$.
- Proof by induction $\frac1{1 \cdot 2} + \frac1{2 \cdot 3} + \frac1{3 \cdot 4} + \cdots + \frac1{n \cdot (n+1)} = \frac{n}{n+1}$
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- Prove elementarily that $\sqrt {(n+1)!} - \sqrt {n!}$ is strictly decreasing
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Yes, the procedure is correct. If you want to write this more like the sort of mathematical proof that would be found in a textbook, you might want to make some tweaks.

For example, the base case could be re-written as follows:

When $n = 4$, we have $2^4 = 16 < 24 = 4!$

Next, the inductive hypothesis and the subsequent manipulations:

Suppose that for $n \geq 4$ we have $2^n < n!$

Thus, $2^{n+1} < 2 \cdot n! < (n+1)!$, where the first inequality follows by multiplying both sides of the inequality in our IH by $2$, and the second follows by observing that $2 < n+1$ when $n \geq 4$.

Therefore, by the Principle of Mathematical Induction, $2^n < n!$ for all integers $n \geq 4$. Q.E.D.

**Note:** I am not making a judgment about whether your write-up or the one I have included here is “better.” I’m only observing that the language and format differ, particularly with regard to proofs that are written in paragraph form (typical of math papers) rather than with a sequence of bullet-points (which is what you had).

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