Prove by induction: $n! \ge 2^{(n-1)}$ for any $n \ge 1$

Question

I need help with this exercise. What I've done so far is prove the exercise when $n=1$. So:

$$n=1$$ $$1!\ge2^{(1-1)}$$ $$1\ge2^0$$ $$1\ge1$$ Which is true

Therefore, now that I assume that the assumption is correct, I want to prove that with $n+1$, it will also be true. So, what I've done now is:

$$P(n) \implies P(n+1) $$

$$n!+n+1\ge 2^{(n-1)}...$$ And my problem is that I do not know how to add the $n+1$ in the right side of the equation, therefore I'm not been able to finish the exercise.

Thank you so much for your help. If something's not very clear, please let me know. I'll try to be clearer next time :)

Answer 1

Assume $n!\ge 2^{n-1} $ for some $n\ge 1$.

$$(n+1)!=(n+1)n!\ge (n+1)2^{n-1} $$

but $n+1\ge 2$ thus

$$(n+1)2^{n-1}\ge 2^{n-1+1} $$ Done!

Answer 2

$P(n+1)$ asserts that $(n+1)!\geqslant2^n$. And your are assuming that $n!\geqslant2^{n-1}$. But then$$(n+1)!=(n+1)\times n!\geqslant2\times2^{n-1}=2^n.$$

Answer 3

For $n=1$ it's obvious.

For $n\geq2$ we obtain: $$n!=n(n-1)...2\geq2\cdot2\cdot...\cdot2=2^{n-1}.$$

Answer 4

Hint Assuming $$n!\ge 2^{(n-1)}$$ Multiply by 2 $$n!2\ge 2^{n}$$ $$(n+1)! \ge n!2\ge 2^{n}$$ For $n+1 \ge 2 \implies n \ge1$ we have that $$(n+1)! \ge 2^{n}$$