Show $\lim_{n \rightarrow \infty} \frac{n}{e^n} = 0$ $$\lim_{n \rightarrow \infty} \frac{n}{e^n}$$
$$= \lim_{n \rightarrow \infty} \exp(\ln(\frac{n}{e^n})) $$
$$= \lim_{n \rightarrow \infty} \exp(\ln(ne^{-n})) $$
$$= \lim_{n \rightarrow \infty} \exp(\ln((n^{\frac{-1}{n}}e)^{-n})) $$
$$= \lim_{n \rightarrow \infty} \exp(-n \ln((n^{\frac{-1}{n}}e))) = 0 $$

Is my argument valid?
Notice, your last step is hard to understand &amp; justify. Although it is easy to apply L'Hospital's rule for $\frac{\infty}{\infty}$ form as follows $$\lim_{n\to \infty}\frac{n}{e^n}=\lim_{n\to \infty}\frac{\underbrace{n}_{\large\to \infty}}{\underbrace{e^n}_{\large \to \infty}}$$ $$=\lim_{n\to \infty}\frac{\frac{d}{dn}(n)}{\frac{d}{dn}(e^n)}$$ $$=\lim_{n\to \infty}\frac{1}{e^n}$$$$=\frac{1}{\infty}=0$$
Yes, it is; although you might want to read a bit about L'Hospital's Rule.