I want to prove that if $0 \le |x| < a$ then: $$ \sum\limits_{n=1}^{\infty} \frac{|x|^{n-1}}{n!} < \sum\limits_{n=1}^{\infty} \frac{a^{n-1}}{n!} $$

I initially attempted to use induction to prove that $\forall n > 1, \frac{|x|^{n-1}}{n!} < \frac{a^{n-1}}{n!}$, but then learned that it is not the appropriate way of proving this.

Am I even correct to assume that the above statement can be proven by showing that $\frac{|x|^{n-1}}{n!} < \frac{a^{n-1}}{n!}$ holds $\forall n > 1$? What is the proper way of approaching this?

Does it suffice to show that $\lim\limits_{n \to \infty} \frac{|x|^{n-1}}{a^{n-1}} < 1$?

Thanks in advance.