# Prove using mathematical induction that for all $n! \ge 2^{n-1}$ [duplicate]

Prove using mathematical induction that for all $$n! \ge 2^{n-1}$$

Base case, p(1), 1! >= 1

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

• n+1 is at least equal to 2.....so expression on the right of your last inequality is larger than 2^n – Alex M. May 27 at 0:23
• The statement can also be proven without using mathematical induction: just observe that $n!$ is the product of $n-1$ integers $\ge 2$ (namely, the integers from $2$ to $n$), and so is $\ge 2^{n-1}$. – Geoffrey Trang May 27 at 0:40
• Does this answer your question? Prove the inequality $n! \geq 2^n$ by induction – Culver Kwan May 27 at 1:27

$$(n+1)! = n!(n+1)\\ (n+1)! \geq 2^{n-1}(n+1)$$
Now for all $$n\geq1$$, $$n+1 \geq 2$$. Therefore, from above, we've shown that
$$(n+1)! \geq 2^{n-1}(2) = 2^n$$
This completes the proof that $$\forall n\geq 1,\quad n! \geq 2^{n-1}$$.