Before reading this please look at the recurrence relation here, and the accepted solution. Upon further inspection, I found another solution to the recurrence relation, namely the reciprocal of: $$\sum_{k=1}^{n-m} \frac{(n-m-k+3)(k+m-3)!}{(m-2)!(k-1)!} + \sum_{k=0}^{m-2} \frac{(2^{m-k})(n-m+k-1)!}{(k)!(n-m-1)!}$$ for $n>m>1$. The powerpoint here might help understand how I found this expression. I am trying to directly convert this expression to (the reciprocal of) the one in the link above. That is, I want to prove for $n>m>1$: $$\sum_{k=1}^{n-m} \frac{(n-m-k+3)(k+m-3)!}{(m-2)!(k-1)!} + \sum_{k=0}^{m-2} \frac{(2^{m-k})(n-m+k-1)!}{(k)!(n-m-1)!} = \sum_{k=0}^{m} \binom{n}{k}$$I am completely stumped - I have only noticed that the factorials in the left hand side form binomial coefficients, and all 3 individual sums having different limits seems to make a potential proof really messy.
My question is, how can I prove that the equation above holds with a method not using the aforementioned problem. I am confident they are the same, as I have thoroughly checked the derivation of the longer expression, proven the shorter expression by simply checking if the recurrence and initial values hold, and comparing both the expressions for a lot of different values for $n, m$, such that $n>m>1$, with a program. Any ideas or hints on proving the equation are welcome. The motivation behind this is to get a useful identity for$\ _2F_1(1, b; c; \frac{1}{2})$, as the left hand side can be written in terms of this.