Climb a step to a prime The title of Numberphile's newest video, 13532385396179, is a fixed point of the following function \$f\$ on the positive integers:

Let \$n\$ be a positive integer. Write the prime factorization in the usual way, e.g. \$60 = 2^2 \cdot 3 \cdot 5\$, in which the primes are written in increasing order, and exponents of 1 are omitted. Then bring exponents down to the line and omit all multiplication signs, obtaining a number \$f(n)\$. [...] for example, \$f(60) = f(2^2 \cdot 3 \cdot 5) = 2235\$.

(The above definition is taken from Problem 5 of Five $1,000 Problems - John H. Conway)
Note that \$f(13532385396179) = f(13 \cdot 53^2 \cdot 3853 \cdot 96179) = 13532385396179\$.
Task
Take a positive composite integer \$n\$ as input, and output \$f(n)\$.
Another example
\$48 = 2^4 \cdot 3\$, so \$f (48) = 243\$.
Testcases
More testcases are available here.
   4 -&gt; 22
   6 -&gt; 23
   8 -&gt; 23
  48 -&gt; 243
  52 -&gt; 2213
  60 -&gt; 2235
 999 -&gt; 3337
9999 -&gt; 3211101
Python, 166 162 159 bytes
You guys are much better. This is what I used! (the algorithm that solved it calls this)

from primefac import*
def c(n):
 x=factorint(n)
 a=''
 for i in range(len(x)):
  l=min(x.keys())
  a+=str(l)
  if x[l]&gt;1:a+=str(x[l])
  x.pop(l)
 return int(a)