Prove $n! > 2^n-1$ for $ n \geq 4$ Prove $n! > 2^n-1$ for $ n \geq 4$
Base Case - $(n=4)$ $\implies$ $24 >15$
Assume true for $n=k$ $\implies$ 
$$k! > 2^k -1$$
$$(k+1)k! > 2^k-1$$
$$(k+1)! > 2^{k}-1$$
$$2(k+1)! > 2^{k+1}-2$$
$$2(k+1)! > (2^{k+1}-1)-1$$
I'm not sure how to continue now..
 A: \begin{align}
(k+1)k! &> (k+1)(2^k-1)\\
&> 2(2^k-1)\\
&> 2^{k+1}-2
\end{align}
Thus $(k+1)!\ge 2^{k+1}-1$, but the left side is even while the right side is odd, so equality is not an option, and $(k+1)!>2^{k+1}-1$.
A: you didn't had a mistake, but the way is not the right approach:
if you assume it is true for $n=k$ to show it is true for $k+1$ you need to show that:$$(k+1)!\ge2^{k+1}-1\\\text{we can do the following: }(k+1)k!\ge(k+1)\times\left(2^{k}-1\right)\gt2\times\left(2^{k}-1\right)$$ because we assumed that $k!\ge2^k-1$ for $k\ge4$ if we show that $k+1\gt2$(which is true) we get $(k+1)!\gt2^{k+1}-2$ because both are even and $(k+1)!\ne2^{k+1}-2$ i can add one to the right side and i get $$(k+1)!\gt2^{k+1}-1$$and you had shown it is true to $k=4$ thus it is true $\forall k\ge4$
A: You are making it more difficult than it needs to be. The $n! > 2^n -1$ is in integers the same as $n! \geq 2^n $. This bound is sharper, and you get rid of the annoying $-1$. Notice that the induction step is then simply $$(n+1)! = (n+1) n! \geq 2\cdot 2^n = 2^{n+1 },$$ where we used $n+1\geq 2$ and the induction hypothesis $n! \geq 2^n$.
