If the question is also applicable to which formulas should be in your cryptographic "**toolbox**" (and that are beautifully simple) I would add [Boltzman's entropy equation][1] (for calculating Entropy which is `s = k(logW)`, but swapped with **[Claude Shannon's interpretation][2]** as it also structurally relates to information theory (and not the decay of gas) and is something that every cryptographer must know how to do, important for combinatorics and security assumption values (passwords, private keys, ciphertext, etc..).

**It's also beautifully simple**, with a few ways to write it:

${log_2(L^N)}$ = **Entropy in bits** (where L is the size of the library and N is the length of the string.) 

Formul ([source][3]) : 

The entropy of a discrete random variabe X is defined as:  ${\\H(X) = E {\lfloor \log
\frac{1}{P_i}
\rceil}
=
\sum_{i}
 P_i \  log
 \frac{1}{P_i}
}$

and where the sum is over the range of X, and ${P_i}$ =
${Pr[x=i]}$

For example, if `X` is a uniform random variable on a string of `r` bits, each ${P_i = \frac{1}{2^r}}$ so that ${log\frac{1}{P^i} = }$`r` thus the expected entropy of `x` is ${H(X)=}$`r`.

(i.e. if x is a binary number where i=2, (base two) and is 256 bits long, where r=256 denoting the length of x, the maximum entropy of x is 256, where h=256). 
   

Put differently, the entropy "H" of a discrete random variable "X" is defined as:  

${\\H(X) = -
\sum_{i=1}^{n}
 P(x_i) \  log_b P(x_i)
}$

**If we equate beauty with its usefulness:** then again, I think every cryptographer should be able to - at a minimum - calculate entropy when dealing with any random length of any text character (number or string) in numerous situations related to cryptographic operations in order to calculate the potential message space and determine the potential maximum theoretical Entropy as bits of security (i.e. 128-bit security, 128 bits of entropy).

Note: the formula doesn't tell us whether the number is random or not or if bits are distributed uniformly, but simply helps calculate the minimum range of numbers an attacker would need to brute force search to guess/find the random number, and since the random number cannot carry more than 1-bit of entropy for every bit in the number, its maximum entropy is equal to that minimum range. So even if a 256-bit number was sourced from a large range of 512-bit numbers, it still only carries a maximum of 256-bits of entropy as it would be sufficient to search all 256-bit numbers (the minimum range) to find it.

(example potentially random 256-bit binary number: `1000110110001000110011010011011101111111010111100111100010111011000010110010010000000110000100111111010111101101011100010101100111010100100100100100001000110000000111001010011111000011001001110110011001101111010001010111000100100001010110011001111111111101`
And the zero-indexed maximum range of 256-bit binary numbers: `1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111`)


  [1]: https://en.wikipedia.org/wiki/Boltzmann%27s_entropy_formula
  [2]: https://en.wikipedia.org/wiki/Entropy_(information_theory)
  [3]: http://pages.cs.wisc.edu/~dieter/Courses/2010s-CS710/Scribes/PDF/lecture21.pdf