Summation with factorial

Question

I want to understand how this step is performed. Can you tell me that how this value of Po is obtained from the first equation.!

$$P_o=\frac{1}{\sum_{k=0}^\infty\left(\frac{\alpha}{u}\right)^k\frac{1}{k!}}$$

$$P_o=e^{-\frac{\alpha}{u}}$$

Answer 1

$$P_o=\frac{1}{\sum_{k=0}^\infty\left(\frac{\alpha}{u}\right)^k.\frac{1}{k!}}$$

Substitute $\frac{\alpha}{u}=x$.

The denominator is $$f(x)=1+x+\frac{x^2}{2!}+\dots$$

Notice that this is the Taylor expansion for $e^x$, and this yields the answer. $$f(x)=e^x$$ $$P_0= \frac{1}{f(x)} = e^{-\frac{\alpha}{u}}$$