Let $ \ \alpha , x \in \mathbb{R} \ $ be such that $ \ \alpha \geq 1 \ $ and $ \ x>0 \ $. Show that $$1+x^{\alpha} < e^{\alpha x}$$ without using the tools of Calculus (beginning on differentiation) or more advanced mathematics.
I have a clue how to solve it using derivatives and the Taylor series, but I can only use basic properties of limits of functions, I can use nothing that comes from derivatives, integrals, infinite series and other things like that. I tried to approach it via the definition of the number $ \ e = \lim_{k \to \infty} \left( 1 + \frac{1}{k} \right)^k \ $ and the Bernoulli's inequality, but I failed.
Any help is appreciated.
EDIT: Like I said in the comments, the definition is $$e^s = \lim_{k \to \infty} \left( 1 + \frac{s}{k} \right)^k \ \ \ , $$ for all $ \ s \in \mathbb{R} \ $.
First, we can start with the inequality $1+x^\alpha \leq (1+x)^\alpha$ for $x>0, \alpha\geq 1$. For $\alpha=1$, this is obvious as equality holds, and otherwise, we can write \begin{align*} 1 &= (1+x) -x & \text{obvious}\\ 1&\leq(1+x)^{\alpha-1}[(1+x)-x] &\text{since } (1+x)^{\alpha-1}>1\\ 1&\leq(1+x)^\alpha - (1+x)^{\alpha-1}x < (1+x)^\alpha-x^\alpha &\text{since } -(1+x)^{\alpha-1}<-x^{\alpha-1}\\ 1+x^\alpha &<(1+x)^\alpha &\text{rearranging} \end{align*} Then, we just have to prove $1+x<e^x$, which can be done by using your limit definition, that $$e^x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n$$ And since we have $$\left(1+\frac{x}{n}\right)^n<\left(1+2\frac{x}{2n}+\frac{x^2}{4n^2}\right)^n=\left(1+\frac{x}{2n}\right)^{2n}$$ We can write $$1+x<\left(1+\frac{x}{2}\right)^2<\cdots \to e^x$$ And so we're done, as we know $1+x$ is the beginning of an increasing sequence which tends to $e^x$!