We find the answer, using basic calculus. The magic $\alpha$ is not $-1$. Consider the function $f(x)=(1+x)^n -(1+nx)$ for $n \gt 1$. Then
$$f'(x)=n((1+x)^{n-1}-1).$$
For $n\gt 1$ and even, the derivative vanishes only at $x=0$. For $n\gt 1$ and odd, the derivative vanishes at $x=-2$ and $x=0$.
The only interesting case is $n$ odd. By the usual increasing/dcreasing analysis, we find that $f(x)\ge 0$ when $x\ge -2$. For any $w\lt -2$, by taking a suitably large $n$, we can make $f(w)\lt 0$.
The part about $f(x)\ge 0$ for $x\ge -2$ can be pushed through by induction.
You wrote:
i understand how to do ordinary induction proofs and i understand that strong induction proofs are the same as ordinary with the exception that you have to show that the theorem holds for all numbers up to and including some n (starting at the base case) then we try and show: theorem holds for $n+1$
No, not at all: in strong induction you assume as your induction hypothesis that the theorem holds for all numbers from the base case up through some $n$ and try to show that it holds for $n+1$; you don’t try to prove the induction hypothesis.
In your example the simple induction hypothesis that the result is true for $n$ is already enough to let you prove that it’s true for $n+1$, so there’s neither need nor reason to use a stronger induction hypothesis. The proof by ordinary induction can be seen as a proof by strong induction in which you simply didn’t use most of the induction hypothesis.
I suggest that you read this question and my answer to it and see whether that clears up some of your confusion; at worst it may help you to pinpoint exactly where you’re having trouble.
Added: Here’s an example of an argument that really does want strong induction. Consider the following solitaire ‘game’. You start with a stack of $n$ pennies. At each move you pick a stack that has at least two pennies in it and split it into two non-empty stacks; your score for that move is the product of the numbers of pennies in the two stacks. Thus, if you split a stack of $10$ pennies into a stack of $3$ and a stack of $7$, you get $3\cdot7=21$ points. The game is over when you have $n$ stacks of one penny each.
Claim: No matter how you play, your total score at the end of the game will be $\frac12n(n-1)$.
If $n=1$, you can’t make any move at all, so your final score is $0=\frac12\cdot1\cdot0$, so the theorem is certainly true for $n=1$. Now suppose that $n>1$ and the theorem is true for all positive integers $m<n$. (This is the strong induction hypothesis.) You make your first move; say that you divide the pile into a pile of $m$ pennies and another pile of $n-m$ pennies, scoring $m(n-m)$ points. You can now think of the rest of the game as splitting into a pair of subgames, one starting with $m$ pennies, the other with $n-m$.
Since $m<n$, by the induction hypothesis you’ll get $\frac12m(m-1)$ points from the first subgame. Similarly, $n-m<n$, so by the induction hypothesis you’ll get $\frac12(n-m)(n-m-1)$ points from the second subgame. (Note that the two subgames really do proceed independently: the piles that you create in one have no influence on what you can do in the other.)
Your total score is therefore going to be
$$m(n-m)+\frac12m(m-1)+\frac12(n-m)(n-m-1)\;,$$
which (after a bit of algebra) simplifies to $\frac12n(n-1)$, as desired, and the result follows by (strong) induction.
Best Answer
What you have is perfectly acceptable. The calculations could be organized a little more neatly:
$$\begin{align*} (1+x)^{k+1}&=(1+x)(1+x)^k\\ &\ge(1+x)(1+kx)\\ &=1+(k+1)x+kx^2\\ &\ge1+(k+1)x\;, \end{align*}$$
since $kx^2\ge 0$. This completes the induction step.