Why is the Keyspace of a Substitution Cipher not 2^26 but 2^88 It is known that a cipher has a keyspace of cryptographic algorithm whose key length is $n$ is given by $2^n$, but the keyspace of the substitution cipher is $2^{88}$ which is an approximation of $26!$. (I am considering an alphabet of 26 letters)

Assuming the key length is 26 (which it should be for the alphabet) Why is it not $2^{26}$?
The key space of a cryptographic algorithm whose key length is $n$ is given by $2^n$


No. There is confusion between:


keyspace (or key space) $\mathcal K$, which is the set of possible keys.
keyspace size (or size of the keyspace) $\|\mathcal K\|$, which is the number of possible keys (an integer).
key length (or key size) in bit, which can be defined as the base-2 logarithm of the keyspace size (not necessarily an integer). If we note $k$ for the key length in bit, the keyspace $\mathcal K$ has $\|\mathcal K\|=2^k$ keys, and $k=\log_2(\|\mathcal K\|)$.
number of symbols used for a representation of a key. If the key is represented as $n$ symbols which each can take $s$ values, then $\|\mathcal K\|\le s^n$ and $k\le n\log_2(s)$. These inequalities turn to equalities when all combinations of symbols yield valid keys (which is typical in modern symmetric ciphers).


This is a slight simplification: often, and always when effective is added before key, equivalent keys (that is keys yielding the same effect on decryption) are aggregated and counted as one. However this distinction is not useful for a mono-alphabetic substitution cipher.

In a mono-alphabetic substitution cipher for an alphabet of 26 letters, with the key represented as a list of ciphertext letters for each plaintext letter in alphabetical order:


The keyspace is the set $\mathcal K$ of all permutations of the alphabet.
The keyspace size is the number of elements in that set, that is the integer $\|\mathcal K\|=26!=403291461126605635584000000$.That's because plaintext letter A can correspond to any of 26 ciphertext letters, plaintext letter B can correspond to any of the 25 ciphertext letters other than the one corresponding to plaintext letter A, plaintext letter C can correspond to any of the 24 ciphertext letters other than the two corresponding to plaintext letters A or B, and so on until plaintext letter Y corresponding to 2 possible ciphertext letters, and Z corresponding to the one remaining ciphertext letter.
The key length in bit is the base-2 logarithm of that: $k=\log_2(26!)=\log(26!)/\log(2)\approx88.38$ bit.
The number of symbols in the key is $n=26$, with $s=26$ possible symbols.



  Assuming the key length is 26, why is (keyspace size) not $2^{26}$ ?


That $2^{26}$ seems to be built as $2^n$, where $n$ is the number of symbols (not key length), when $2$ would be for the number of possible values taken by a single bit. The discrepancy makes $2^{26}$ irrelevant.

A relevant number is $s^n=26^{26}=6156119580207157310796674288400203776=2^{\approx122.21}$ : that's the number of character strings of $n=26$ characters each chosen among $s=26$ characters. But this is much larger than the keyspace, because the key of a mono-alphabetic substitution cipher must use different letters (otherwise decryption would be ambiguous).