Why do we need Euler's totient function $\varphi(N)$ in RSA? After we calculated $N = p * q$, we calculate $\varphi(N)$ and use it later to determine $e$ (PR) and $d$ (PU). But why?
For decryption and encryption, we only use $N$ and don't need $\varphi(N)$. So why can't we find $e$ and $d$ without Euler's totient function?
I know that $\varphi(N)$ is giving me how many integers are coprime to $N$, and if $N$ is a prime then it would be $\varphi(N) = N - 1$. But why is this useful? Or rather why is it a must for determining $e$ and $d$ in RSA?
Edit:
And why does $e$ need to be smaller than $\varphi(N)$?
What we really need is a number $\lambda$ satisfying $x^{\lambda+1} \equiv x \pmod n$ for all integers $x$ (which, by induction, then implies that $x^{k\lambda+1} \equiv x \pmod n$ for any $k$).

Given such a $\lambda$, and an arbitrary encryption exponent $e$ which is coprime to it, we can then find the multiplicative inverse of $e$ modulo $\lambda$, i.e. a number $d$ such that $ed \equiv 1 \pmod \lambda$, or in other words, $ed = k\lambda + 1$ for some integer $k$.  Such $e$ and $d$ then satisfy $$(x^e)^d = x^{ed} = x^{k\lambda+1} \equiv x \pmod n,$$ meaning that, if we encrypt a number by raising it to the $e$-th power modulo $n$, we can recover the original number by raising the result to the $d$-th power and again reducing it modulo $n$.  This is what we need for RSA encryption and decryption to work correctly.

The smallest such number $\lambda$ is given by the Carmichael totient function, which, for a product $n = pq$ of two primes, is $$\lambda(pq) = \operatorname{lcm}(p-1, q-1)$$ where $\operatorname{lcm}(p-1,q-1)$ denotes the least common multiple of $p-1$ and $q-1$.  However, since we don't necessarily need the smallest such number, it's also possible to use the Euler totient function $$\varphi(pq) = (p-1)(q-1)$$ which is, by definition, always a multiple of $\lambda$.

I'm not aware of any particular reason for preferring $\varphi$ over $\lambda$, except that it's slightly easier to compute $\varphi$ and to explain why it has the necessary property $x^{\varphi+1} \equiv x \pmod n$, which may be why introductory texts tend to prefer it.  I do believe, however, that actual practical RSA implementations (insofar as they explicitly compute the decryption exponent at all, rather than e.g. using the Chinese remainder theorem) generally use $\lambda$ rather than $\varphi$, since doing so yields the smallest possible decryption exponent $d$.



Addendum: The reason why $\lambda$ (and $\varphi$) satisfies $x^{\lambda+1} \equiv x \pmod n$ is basically Fermat's little theorem, which says that, for any prime $p$ and any integer $x$, $$x^p \equiv x \pmod p.$$

This can be easily generalized to show that, for any multiple $k\lambda(p)$ of $\lambda(p) = p-1$, $$x^{k\lambda(p)+1} \equiv x^{\lambda(p)+1} = x^p \equiv x \pmod p.$$

In particular, since $\lambda(pq) = \operatorname{lcm}(p-1,q-1)$ is, by definition, a multiple (and in fact, the smallest common multiple) of both $\lambda(p) = p-1$ and $\lambda(q) = q-1$, it follows that $x^{\lambda(pq)+1} \equiv x$ modulo both $p$ and $q$, and therefore, also modulo $pq = n$.

Of course, since $\varphi(pq) = (p-1)(q-1)$ is also a multiple of both $p-1$ and $q-1$, it also has the same property, as does any other common multiple of those numbers.

The fact that $\varphi(n)$ also happens to be the order of the multiplicative group modulo $n = pq$ is basically a red herring; there's nothing special about $\varphi(n)$ among all the other multiples of $\lambda(n)$ as far as RSA is concerned.  What's more relevant is that $\lambda(n)$ is the exponent of this group, which is essentially another way of stating the crucial property that $x^{\lambda(n)+1} \equiv x \pmod n$ for all $x$.