I just started learning lambda calculus and I understood most of it but i was thinking that why do we define $n=\lambda f.\lambda x.f^n(x)$ instead of $n=\lambda f.f^n$? I think it would be more simple in that way?
$$0\equiv\lambda f.\;\lambda x.x$$ $$1\equiv\lambda f.\;\lambda x.\; f(x)$$ $$2\equiv\lambda f.\;\lambda x.\; f(f(x))$$ $$3\equiv\lambda f.\;\lambda x.\; f(f(f(x)))$$ in normal lambda calculus but I think it would be better if $$0\equiv\lambda x.\,x$$ $$1\equiv\lambda x.\,xx$$ $$2\equiv\lambda x.\,xxx$$ $$3\equiv\lambda x.\,xxxx$$
I can define the usual successor function and the addition function with this definition but I have not tried to implement multiplication. But it is quite similar to addition and I will be able to do it. It seems that this definition is also valid.
Please provide some reason for using the standard definition?