About the combinatorial question: the proof follows from the identity $$ {2k-1 \choose i-1} + {2k-1 \choose i} = {2k \choose i} $$ Thus $$ \sum_{i=0}^{k-1} {2k\choose i} p^{2k-i} (1-p)^i = \sum_{i=0}^{k-1} {2k-1 \choose i} p^{2k-i} (1-p)^i + \sum_{i=0}^{k-1} {2k-1 \choose i-1} p^{2k-i} (1-p)^i $$ which implies $$ \sum_{i=0}^{k-1} {2k\choose i} p^{2k-i} (1-p)^i = \sum_{i=0}^{k-2} {2k-1 \choose i} p^{2k-1-i} (1-p)^i + {2k-1 \choose k-1} p^{k+1} (1-p)^{k-1} $$ and $$ \frac{1}{2}{2k\choose k} = {2k-1\choose k-1} $$ implying that $$ {2k-1 \choose k-1} p^{k+1} (1-p)^{k-1} + \frac{1}{2}{2k\choose k} p^{k} (1-p)^{k} = {2k-1\choose k-1 } p^{k} (1-p)^{k-1} $$ which establishes your identity.
About your "Edit" question, I think you mean the theorem in complex calculus that states that, if an analytic function is constant over an interval, it is constant everywhere.