Multiprime RSA proof of correctness

[beaute]

I would like to prove the correctness of multiprime RSA. I have looked up online resources but I am not sure I fully understand what needs to be done.

The Multiprime version of RSA is described as follows:

We let r to be a fixed positive integer where r >= 2,
and p[sub:1zooc5ut]1[/sub:1zooc5ut],...,p[sub:1zooc5ut]r[/sub:1zooc5ut] are distinct odd prime numbers.
we then set n = ?[sub:1zooc5ut]i=1[/sub:1zooc5ut][sup:1zooc5ut]r[/sup:1zooc5ut]p[sub:1zooc5ut]i[/sub:1zooc5ut]. n is the modulus,
we also have ?(n) = ?[sub:1zooc5ut]i=1[/sub:1zooc5ut][sup:1zooc5ut]r[/sup:1zooc5ut](p[sub:1zooc5ut]i[/sub:1zooc5ut]-1)
we have e which is an integer with 1 < e < ?(n) such that gcd(e,?(n))=1.
there is a unique d with 1<d<?(n) such that ed=1 mod ?(n). we also define x??/n?.

finally we have the encryption function e(x) = x[sup:1zooc5ut]e[/sup:1zooc5ut] and
the decryption function d(x)=x[sup:1zooc5ut]d[/sup:1zooc5ut]

I need to prove that for all x??/n?, i have d(e(x))=x.

I started developing the solution based on the proof of correctness of the original rsa. here is what i have done so far:

d(e(x)) = (x[sup:1zooc5ut]e[/sup:1zooc5ut] mod n)[sup:1zooc5ut]d[/sup:1zooc5ut] mod n
? x[sup:1zooc5ut]ed[/sup:1zooc5ut] mod n

from ed ? 1 (mod ?(n)) i have k such that ed = k?(n)+1
so i have x[sup:1zooc5ut]k?(n)+1[/sup:1zooc5ut] (mod n)

i then apply fermat's little theorem since i know x and n are coprime, so
x[sup:1zooc5ut]?(n)[/sup:1zooc5ut] ? 1 (mod n)
so x[sup:1zooc5ut]ed[/sup:1zooc5ut] ? x[sup:1zooc5ut]k?(n)+1[/sup:1zooc5ut]
?x[sup:1zooc5ut]k?(n)[/sup:1zooc5ut] . x[sup:1zooc5ut]1[/sup:1zooc5ut]
? 1 (mod n)
? x

this is where i get lost, i have many prime numbers because ?(n) is the product of p[sub:1zooc5ut]i[/sub:1zooc5ut]-1. I know I have to use the chinese remainder theorem, i'm just not sure how.