RSA algorithm: Must $e$ be less than $\varphi(N)$? I understand why $e$ must be co-prime to the totient of $N$, but I see in many explanations of RSA that $e$ should be less than $\varphi(N)$. Is there a reason for this? Is it for ease of a small number or is there some law in play here?
First, the Euler's theorem

if $n$ and $a$ are coprime positive integers then
$$a^{\varphi(n)} \equiv 1 \pmod n$$ where $\varphi(n)$ is Euler's totient function.

Take $e&gt; \varphi(n)$. Then, there exists $e' &lt; \varphi(n)$ such that: $$e \equiv e' \pmod {\varphi(n)}$$
Then, we can write $$e = e' + k \cdot \varphi(n).$$
This $e'$ is the smaller representation of $e$ that does the same job.
$$a^{e} \equiv  a^{e' + k \cdot \varphi(n)} \equiv a^{e'} a^{k \cdot \varphi(n)}  \equiv   a^{e'} (a^{\varphi(n)})^k  \equiv a^{e'} (1)^k  \equiv a^{e'} \pmod n$$ Therefore, there is no need for $e &gt; \varphi(n)$.
We choose $e$ small so that the encryption can take a small amount. This is what we can control. We don't want the private exponent $d$ small, since it is insecure. If we take a random $e$, then we will not have the benefit.
Commonly used $e$'s are $\{3, 5, 17, 257\text{ or }65537\}$ and there is no danger even with $e=3$, as long as correct padding is used.
For encryption: Use Optimal Asymmetric Encryption Padding (OAEP) (RSA-OAEP) or RSA PKCS#1 v1.5 padding, the former is easier to implement, the later had many attacks due to improper implementations.
For signatures: Use Probabilistic Signature Scheme (PSS) and known as RSA-PSS.

Note 1: We first choose the $e$, then we choose the primes, not the reverse, since we may not get $\gcd(e,\varphi(n)) =1$  for all $n = p \cdot q$ is product of two distinct primes $p$ and $q$.
Note 2: We use Carmichael lambda $\lambda$ instead of $\varphi$, since it can produce smaller $d$s, since $\lambda(n)∣\varphi(n)$