Proper Way to Solve Mathematical Induction? Say I have $2^n +1 < n! -n$ for all $n \ge 4$, and $n$ is an integer.
My inductive steps says, consider a $k$ that is a arbitrary integer, assuming $P(k)$.
Thus,
$$\begin{align} 2^{k+1} +1 = 2^k \cdot 2 +1 \\ 2^k\cdot 2+1\overset{\mathrm{IH}}{<} 2(k!-k) < ((k+1)!-k) \\\text{Therefore, we have proven this my mathematical induction.}
 \end{align}$$
I am not too sure on how to use < in mathematical induction. I based this off an example I saw, and wondering it is valid, and maybe some calcification on what this works (if it does)...
 A: Assuming $P(k)$ you have to show $P(k+1)$. In other words, assuming that $$2^k + 1 < k! - k,\tag{$P(k)$}$$ you have to show that
$$2^{k+1} + 1 < (k+1)! - (k+1). \tag{$P(k+1)$}$$
Your inductive hypothesis $P(k)$ can be rewritten as
$$2^k < k! - k - 1\tag{*}\label{*}.$$
Thus
\begin{align}
2^{k+1}+1
&= 2\cdot 2^k + 1\\
&\stackrel{\eqref{*}}{<}2(k!-k-1)+1\\
&=2k! - 2k -1\\
&<2k!-(k+1)\\
&<(k+1)!-(k+1).
\end{align}
A: In general, mathematical induction on the statement $P(k)$ provides a proof strategy requiring the following steps:

*

*Prove the base case: $P(k=0)$ is true


*Assume the inductive hypothesis: $P(k)$ is true for some $k \in \mathbb{N}.$


*Show that as a result of the above: $P(k+1)$ is true.


*The proof is complete for all $k$.
In this case, your $P(k)$ is the statement $2^k + 1 < k! - k \ \forall k \geq 4 $.
Let us show $P(k=4)$ holds (the smallest value for the inequality): $2^4 + 1 = 17 < 4! - 4 = 24-4=20$ is clearly satisfied.
Now let us assume $2^k + 1 < k! - k$ for some $k \geq 4$ (since we've already proven the case $k=4$).
We want to use this to prove $2^{k+1}+ 1 < (k+1)! - (k+1)$
As you've mentioned, $2^{k+1}=2*2^{k}$. Recalling the inductive step above,
$2*2^{k} < 2*(k!-k - 1).$ Certainly $2k! - 2k - 2 = 2k! - 2*(k+1) < (k+1)! - (k+1)$.
This is because $2k! < (k+1)! = (k+1)k!\  \forall k > 1$.
Now, chaining together the last few inequalities, we arrive at $2*2^k=2^{k+1} < (k+1)! - (k+1)$
