Induction Proof Verification via Binomial Theorem I am studying from a text with an overview of discrete mathematics in the beginning. The problem in question is to prove by induction that
$(1+x)^{k+1}$ $\geq$ $1+nx$ for all naturals, where $x > 1$
I proved this by induction in a similar vein as the solution in the back of the book. The following 'proof' in this post was my first attempt, but I was unsure about its validity. Specifically in the induction step (proving that $n+1$ is true) I am afraid that I am making the assumption that it is true before showing that it is true.
Proof
As a base case, take $n=1$. In this case the inequality holds. Now suppose this is true up to the $k$th natural number. We will now show that this is true for $k+1$.
$(1+x)^{k+1}$ $\geq$ $(1+(1+k)x)$
By the binomial theorem the left and right side give
$x^{0}$ + 
${k+1\choose 1}x$ + ... + $x^{k+1}$ $\geq$ $1+x(k+1)$. 
After evaluating the left hand side we obtain
$1$ + 
$(k+1)x$ + ... + $x^{k+1}$ $\geq$ $1+(k+1)x$ . 
Subtract the right hand side from the first two terms on the left and we have
${k+1\choose 2}x^{2}$ + ... + $x^{k+1}$ $\geq$ $0$ . 
Because the left hand side is the sum of positive terms* it is certainly greater than or equal to zero.
QED
*I suppose this is only true if for $x>-1$ where $x$ is a natural. Should x be a negative real number less than zero but greater than negative one this would not hold. Thoughts?
An evaluation of whether or not this is otherwise a valid proof by induction would be greatly appreciated. I'd be grateful to hear any other criticisms of my mathematical writing in general.
Edit: thanks to all who proved this in the comments. I actually did it that way as well. The real reason for the post was because I knew I was making a logical error in how I tried to prove via the binomial theorem.

You needn't invoke the Binomial theorem, and your proof is not inductive.
The induction hypothesis is
$$(1+x)^n\ge 1+nx$$
and you need to show that this implies 
$$(1+x)^{n+1}\ge 1+(n+1)x.$$
Indeed,
$$(1+x)^{n+1}=(1+x)^n(1+x)\ge^1 (1+nx)(1+x)\ge^2 1+x+nx=1+(n+1)x.$$
$^1$ by the induction hypothesis.
$^2$ because we drop the term $x^2$.
As the statement is true for $n=1$, it is true for all $n$.
