Recurrence $a_j = \frac{n}{n-j} + \frac{j}{n-j} a_{j-1}$ Let $n\in \mathbb{N}$. How to solve the recurrence
$$a_j = \frac{n}{n-j} + \frac{j}{n-j} a_{j-1}$$
for $1\leq j <n$, and $a_0=1$?
I calculated it for some $n$s:

$n=2: [1, 3]$
$n=3: [1, 2, 7]$
$n=4: [1, \frac{5}{3}, \frac{11}{3}, 15]$
$n=5: [1, \frac{3}{2}, \frac{8}{3}, \frac{13}{2}, 31]$
$n=6: [1, \frac{7}{5}, \frac{11}{5}, \frac{21}{5}, \frac{57}{5}, 63]$
$n=7: [1, \frac{4}{3}, \frac{29}{15}, \frac{16}{5}, \frac{33}{5}, 20, 127]$

It looks like $a_{n-1} = 2^n - 1$.
I tried to use generating function $A(x) = \sum_{k=0}^{n-1} a_k x^k$ and got
$$(n-x)A(x) - 2xA'(x) = n\frac{1-x^n}{1-x}-(n-1)a_{n-1}x^{n-1} - a_{n-1}x^n$$
but don't know how to solve this or if it's correct (it's highly probable I made a mistake somewhere).

From the recurrence relation, it is easy to check that
$$ \binom{n-1}{j} a_j - \binom{n-1}{j-1}a_{j-1} = \binom{n}{j}. $$
Therefore
$$ a_j = \frac{\sum_{k=0}^{j} \binom{n}{k}}{\binom{n-1}{j}}. $$

Addendum. By a different method, one can also prove that
$$ a_j = n \int_{0}^{1} (1-u)^{n-1-j} (1+u)^j \, du = \sum_{k=0}^{j} \binom{j}{k} (-1)^{j-k}2^k \frac{n}{n-k}. $$
