Proof.
We have
with each
![$ p_i$](img115.png)
prime.
Suppose that
is another expression of
![$ n$](img102.png)
as a product of primes.
Since
Euclid's theorem implies that
![$ p_1 = q_1$](img118.png)
or
![$ p_1 \mid q_2\cdots q_m$](img119.png)
. By induction, we see that
![$ p_1 = q_i$](img120.png)
for some
![$ i$](img121.png)
.
Now cancel
and
, and repeat the above argument. Eventually,
we find that, up to order, the two factorizations are the same.