I was trying to understand cryptography from the book Information Security by Mark Stamp and I am confused by something.
Suppose that Alice has a secure block cipher, but the cipher only uses an 8-bit key. To make this cipher "more secure," Alice generates a random 64-bit key K, and iterates the cipher eight times, that is, she encrypts the plaintext P according to the rule
$$C = E(E(E(E(E(E(E(E(P, K_0), K_1), K_2), K_3), K_4), K_5), K_6), K_7)$$
where $K_0,...,K_7$ are the bytes of the 64-bit key $K$.
- Assuming known plaintext is available, how much work is required to determine the key $K_1$?
- Assuming a ciphertext-only attack, how much work is required to break this encryption scheme?
Here I believe that the author is trying to explain the meet in the middle attack, but how does this attribute to it? As far as I have understood things, by using a MIM attack, the answer will be $2^7$ for each block do is $2^7*8$, and ciphertext I believe there are no ciphertext shortcuts so is should be $2^{64}$?
Am I right or I am thinking in the wrong direction?
Is encrypting 8 times with 8-bit key beneficial?