I have $\sum_{k=0}^{p-1}\frac{(p-1)!}{(p-1-k)!k!}$. Whereby $p$ is prime.
how can I say that the fraction's result will be an integer?
I tried it with numbers and yes it works. So I tried two approaches:
My first try was to write the factorial as a product,which means, $\frac{\prod_{k=1}^{p-1}k}{\prod_{i=1}^{p-1-k}i\prod_{j=1}^{k}j}$ but I didn't get much further and the second one was to try and use some properties of the factorial and I wrote it as: $\frac{1!(p-2)!(p-1)}{k!(p-1-k)(p-2-k)!}$.
In both cases I tried to expand them and stuff but didn't come to any conclusion. If I expand them I do understand why it must work, but I can't put it in some mathematical decent way.
I would like to solve it alone but I would be thankful for some tip, or maybe for some other property that I am missing, so that then I can continue it.
Thanks in advance
Annalisa
This is the binomial expansion $$(a+b)^n = \sum\limits_{k=0}^n{n \choose k}a^kb^{n-k}$$ with $n=p-1$ and $a=b=1$. (Recall the symbol ${n\choose k} = \frac{n!}{k!(n-k)!}$ is an integer when $n$ and $k$ are, as it is the number of combinations of $n$ objects taken $k$ at a time.)
So your sum is $2^{p-1}$.