well....you can prove it like this..

we can define the exponation of cardinals like that:

if

are cardinals of two sets

, then

is the cardinal of the set of all functions mappings

to

.

Let the cardinal of the natural numbers to be

, and the cardinal of real numbers to be

.

Then let

be set of the all finite subsets of a continuum.

For every

,since

is finite, there is a bijection

from

(meaning the set

)to

, thus every

can be written as

. But we can also see

as mapping from

to

, letting all other natural number map to a strange element

different from all real number. Thus now we have mapped all

to some function from

to

(which have the same cardinality as

).And clearly this mapping is one-to-one.

Clearly we have

. From above we also have

.

But

So

, it have the cardinality of a continuum.