I have to prove this preposition by mathematical induction:

$ \left(x^{k}+1\right)<\left(x+1\right)^{k} \quad \forall n\geq 2 \quad \text{and}\quad x>0 \quad$ $ \quad n \in \mathbb{N}$

I started the prove with $ n=2$ :

$ \left(x^{2}+1\right)<\left(x+1\right)^{2}$

$ x^{2}+1<x^{2}+2x+1$

We see that;

$ x^{2}+1-x^{2}-1<2x$

$ 0<2x$

Then

$ x>0$

And this one carries out for $ n=2$

Now for $ \quad n=k \quad$ (Hypothesis)

$ \left(x^{k}+1\right)<\left(x+1\right)^{k}$

We have

$ \displaystyle x^{k}<\left(x+1\right)^{k}-1\ldots \quad (1)$

Then, we must prove for $ \quad n= k+1 \quad$ (Thesis):

$ x^{k+1}+1<\left(x+1\right)^{k+1}$

We develop before expression as:

$ x^{k+1}<\left(x+1\right)^{k+1}-1\ldots \quad (2)$

According to the steps of mathematical induction, the next stpe would be use the hypothesis $ (1)$ to prove thesis $ (2)$ . It’s in here when I hesitate if the next one that I am going to write is correct:

**First way:**

We multiply hypothesis $ (1)$ by $ \left(x+1\right)$ and we have:

$ x^{k}\left(x+1\right)<\left[\left(x+1\right)^{k}-1\right]\left(x+1\right)$

$ x^{k}\left(x+1\right)<\left(x+1\right)^{k+1}-\left(x+1\right)$

Last expression divided by $ \left(x+1\right)$ we have again the expression $ (1)$ :

$ \displaystyle \frac{x^{k}\left(x+1\right)<\left(x+1\right)^{k+1}-\left(x+1\right)}{\left(x+1\right)}$

$ x^{k}<\left(x+1\right)^{k}-1$

**Second way:**

If we multiply $ (2)$ by $ x$ we have:

$ xx^{k}<x\left[\left(x+1\right)^{k}-1\right]$

$ x^{k+1}<x\left(x+1\right)^{k}-x$

And if we again divided last expression by $ x$ , we arrive at the same result

$ \displaystyle \frac{x^{k+1}<x\left(x+1\right)^{k}-x}{x}$

$ x^{k}<\left(x+1\right)^{k}-1$

I do not find another way to prove this demonstration, another way to solve the problem is using Newton’s theorem binomial coeficients, but the prove lies in the technical using of mathematical induction. If someone can help me, I will be very grateful with him/her! Thanks -VĂctor Hugo-