Placing Balls in Urns and Expected values

This is the ole let's place balls in an urn problem (statiscians love
placing balls in urns :p) and I usually like these types of problems.
But this problem is different from what I've ever experienced.

Suppose that n balls are randomly placed in n urns in such a way that
each ball is equally to go into each urn. What are the expected number
of empty urns.

I thought I'd try using the expectation formula, so note to myself...
E(X) = summation of nPx(X)

Thus: n=number of balls and urns (n = 1,2,3,...n)
p = 1/n

So I get: E(X) = (1*1/n) + ((2*(1/n)) + ((3*(1/n)) + ((n-1)*(1/n)) +
(n*(1/n))
= n!/n = (n-1)!

I peek into the back of my book for the answer to this question and
it's not THAT answer.
The answer is: n(1-(1/n))^n

I try to backwards derive how they came to this step since I know the
answer, I start from there. I substitute known values making
k = 0 urns are filled
p=1/n:

B(n,0)=C(n,0) * p^0 * (1-1/n)^(n-0) = 1*1*(1-(1/n)^n)

Eureka, there's some of my answer ..once I simplify it some, it
condenses to the book's answer partially. I'm not sure where that
extra n came from. So I scratch that part out.

So I rethink this again and came up with E(X) = n(1-(1/n))

Agggh. I'm realizing the expection formula has no power to it. I'm
going in circles in my head. :(

Any insight to this is helpful,
Thanks in advance,
Taria the Weary

.

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