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just tell me please if my idea correct is

$$\exp(x) = \sum_{k=0}^{n} \frac{x^k}{k!} > \frac{x^n}{n!} $$

let $x > R= (n+1)!$ $\Rightarrow $ $$\frac{x^n}{n!}>\frac{R^n}{n!} >\frac{R}{n!}>\frac{(n+1)!}{n!}= n+1 $$

let $n_{0} \in \mathbb{N}$ so that n>$n_{0}$ for a big $n_{0}$

for all M >0 let s choose R = M!

then for all $x > R =(n+1)!$ there is a $M =(n+1)$ so that $\exp(x) > M$

$ \Rightarrow \lim_{x \to \infty}\exp(x) = \infty $

is that correct ?

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    More simply, $e^x > x$ for all $x > 0$. – 2017-02-01
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    @AntonioVargas yeah i knw that , but can you tell me if it is correct ? plz – 2017-02-01
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    I have some trouble with your proof. The definition of $\lim_\limits{x\to \infty} \exp(x) = \infty$ is $\forall M, \exists R, \forall x > R, \exp(x) > M$. But you proved that $\exists R, \forall x > R, \exists M, \exp(x) > M$. Or did I miss something? – 2017-02-01
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    yeah sorry i mean for all x , not for all R – 2017-02-01
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    @Mohbenay: Now, you're defining the value of $R$ twice. $x > R = (n+1)!$ and $R = M!$. Also, you should justify (imo) why $R^n/n! > R/n!$, also it should be $\exp(x) = \sum_{k=0}^{\infty} \frac{x^k}{k!} > \frac{x^n}{n!}$ for some $n$. – 2017-02-01
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    But @Antonio Vargas's suggestion is quite simpler, frankly. – 2017-02-01

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