Proving things is not my forte. Stumbled upon following identity:
$$\binom{n+k+1}{k}>\sum_{i=1}^k \binom{\alpha_i}{i}$$ For
$\alpha_i<\alpha_j $ for $i<j$
$0\leq \alpha_i \leq n+k$
Also $\binom{a}{i}=0$ for $a<i$
Similar to the hockey-stick theorem. Not sure how to apply mathematical induction here.
Edit:
The right hand side of the inequality gets maximum value for $\alpha_1=n+1$ and $\alpha_k=n+k$. Therefore
$$\binom{n+k+1}{k} > \sum_{i=1}^k \binom{n+i}{i} $$ comparing with hockey-stick theorem \begin{eqnarray} \binom{n+k+1}{k} &=& \sum_{i=0}^k \binom{n+i}{i} \\ &=&\sum_{i=1}^k \binom{n+i}{i} + 1 \\ &>& \sum_{i=1}^k \binom{n+i}{i}\\ &>& \sum_{i=1}^k \binom{\alpha_i}{i} \end{eqnarray} Q.E.D
Thanks to CuddlyCuttlefish for suggesting maximization of R.H.S of inequality.