Show that there are $\binom{n-1}{k-1}$ ways to describe $n$ as a sum of $k$ positive, ordered parts.
I know that there are $n-k+1$ options for the first part of the sum. And the first choice affects the choice for the snd part and so on. I have also tried a proof by induction, assuming that there are $\binom{n-1}{k-1}$ ways to describe $n$ as a $k$-part sum, we pick one possibility and add 1 to one of the parts and we get $n+1$ That means there would be $k\binom{n-1}{k-1}$ ways to describe $n+1$ but there are doubles such that $(m_1,\dots,m_k)\ne(m'_1,\dots,m'_k)$ but $(m_1,\dots,m_i+1,\dots,m_k) = (m'_1,\dots,m'_j + 1,\dots,m'_k)$ with $i\ne j$ and $m_1 + \dots+ m_k = n = m'_1 +\dots+ m'_k$ and $m'_1 +\dots+ m'_j + 1 +\dots+ m'_k = m_1 +\dots + m_i + 1 +\dots+ m_k = n + 1$.
It would be good if we could show somehow that there are $\binom nk$ ways to describe $n+1$.
These are my ideas maybe I am completely wrong with my approaches if so please tell me, any hint would be appreciated.
You are, indeed, completely missing a direct solution. Picture $n$ stars and $k-1$ bars which go into the $n-1$ gaps between the stars, at most one bar per gap. Then the $\binom{n-1}{k-1}$ ways to put the bars in the gaps are in bijection with the number of ways to describe $n$ as a sum in $k$ (ordered, positive) parts; I leave it to you to show this bijection.