In an attempt to compute $\sum_{k=0}^{n}{(-1)^k {n\choose k}}$ the author put $x=-1$ in the formula $(x+1)^n=\sum_{k=0}^{n}{ {n\choose k}x^k }$
Then he wrote that we have $$0^n = \begin{cases} 0 &\mbox{if } n \in \mathbb N^* \\ 1 & \mbox{if } n =0 \end{cases} $$
How is that done and is the function $0^n$ defined at all? I mean my understanding is that $0^n=e^{n\log0}$ but $\log 0$ is not defined at all. Thank you for your help.
On
The problem is that $x^y$ has a discontinuity at $(0, 0)$. If you take $x = 0$ and $y > 0$, it is $0$; for $x > 0$ and $y = 0$ it is $1$. No way around it. Just be careful.
Now, you can write a polynomial (and power series) as:
$$ a_0 + a_1 z + a_2 z^2 + \dotsb $$
This can be written:
$$ \sum_{n \ge 0} a_n z^n $$
as then the term $a_0 z^0 = a_0$ works out nicely unless $z = 0$. To avoid awkward formulæ, in such kinds of uses the convention that $0^0 = 1$ is adopted.
So, one take is that $0^n = 0$ unless $n = 0$, when it is $1$. Or you could say $0^0 = 0$ to make it continuous on the exponent. Or bite the bullet and declare it isn't defined at all. Take you pick, according to convenience.
It is just a matter of definitions. For a polynomial $p(x)\neq 0$ (in general, for an element $a\neq 0$ of some ring) we have $p(x)^0=1$, so that the relation $p(x)^mp(x)^n=p(x)^{m+n}$ holds true for $m,n\in\mathbb{N}$.