This problem comes from the 1975 IMO Q4. It is stated as follows:
Let $f(n)$ be the digit sum of the integer $n$. ($f(57)=12, f(63)=9$,...). Let $N$=$4444^{4444}$. Find $f(f(n))$.
Proof: Well, consider this (mod $9$). i.e., first lets find a pattern: $4444$=$7$(mod $9$), $4444^2$=$7^2$=$4$(mod $9$), $4444^3$=$7^27$=$1$(mod $9$). Then notice that $4444$=$3(1481)$+$1$, which would then imply $f(4444^{4444})$=$f(f(4444^{4444}))$=$4444^{4444}$(mod $9$), which is $7$. So $f(f(N))$=$7$.
Is my solution correct? Namely that of $7$?
