How would I go about proving that $n! \ge 2^{n-1}\ \forall n \ge 1$? The base case makes sense to me, but when I do the inductive step, I go here using the inductive step:
$$ n+ 1 = k+1 $$ $$ (k+1)! \ge 2^{(k+1)-1} $$ $$ (k+1)! = (k+1)\cdot k!$$ $$\ 2^{k-1+1} = 2^k = 2^{k-1} \cdot 2 $$ $$(k+1) \cdot k! \ge 2^{k-1} \cdot 2 $$ $$ \therefore (n+1) \cdot n! \ge 2^{n-1} \cdot 2$$
Is this a viable proof? If not, what do I need to do differently in order to make it correct? Have I not gone far enough?
Assuming your result holds for $n=k$, i.e. $$k! \geqslant 2^{k-1},$$ you then have $$(k+1)! = (k+1) \times k! \geqslant (k+1) \times 2^{k-1} \geqslant 2^k,$$ where the first equality is the definition of the factorial, the second is the case $n=k$, and the third is true because $k\geqslant1$.