I have the following summation:
$$\sum_{k=0}^\infty(1-e^x)^k=\sum_{k=0}^\infty\sum_{j=0}^k\binom{k}{j}(-1)^je^{jx}$$ Then $$e^{jx}=\sum_{i=0}^\infty j^i\frac{x^i}{i!}$$ So, $$\sum_{k=0}^\infty\sum_{j=0}^k\binom{k}{j}(-1)^j\sum_{i=0}^\infty j^i\frac{x^i}{i!}=\sum_{k=0}^\infty\sum_{j=0}^k\sum_{i=0}^\infty\binom{k}{j}(-1)^j j^i\frac{x^i}{i!}$$ However, when I run the program, my first term with have $j^i=0^0$ I don't feel like any of the math is wrong above, so what is happening?
In this case, $0^0=1$ for it is what happens for $n=0$ term when $x=0$ in $e^x=\sum\limits_{n=0}^{\infty}\dfrac{x^n}{n!}$.