Pandas groupby count unique non cumulative

Question

I have this toy data set

df=pd.DataFrame({'user':['John','Steve','Steve','Steve','Jane','Jane','Jane','Jane','Alice','Alice','Alice'],
                  'days':[1,1,2,3,1,2,3,4,1,2,3]})

yielding

user  days
John    1
Steve   1
Steve   2
Steve   3
Jane    1
Jane    2
Jane    3
Jane    4
Alice   1
Alice   2
Alice   3

I wish to count the exact number of user with only 1 days, only 2 and only 3 days.

Desired output

user   days_count
1           1
3           2
4           1

I have tried code from this answer and from this answer, but non yielded above (or similar result)

Answer 1

Looks like a double value_counts:

df['user'].value_counts(sort=False).value_counts(sort=False)

Output:

count
1    1
3    2
4    1
Name: count, dtype: int64

For the exact format:

out = (df['user'].value_counts(sort=False)
       .value_counts(sort=False)
       .rename_axis('user')
       .reset_index(name='days_count')
      )

Or:

from collections import Counter

c = Counter(Counter(df['user']).values())

out = pd.DataFrame({'user': c.keys(), 'day_count': c.values()})

Output:

   user  days_count
0     1           1
1     3           2
2     4           1