RSA: least significant bit position in key How can I show that $C' = 2^e·C^* \bmod N$ when it is decrypted will add zero bits in the least significant bit position of the key?

I am not sure if the cause is multiplying by a ciphertext which would lead the key to be multiplied by it as well.
Because (if $C=P^e$ where $P$ is the plain text): $2^e C= (2P)^c \pmod{N}$ so multiplying the ciphertext by $2^e$ modifies the plain text by a multiplication of $2$, which is a bitshift to the left of $1$, so we get 1 new zero bit at the end. To get $m$ zero bits, we need to multiply by $2^{em}$ (modulo $N$), of course.