one sum can be carried out: exchanging the order of summation, $$T(N,K)=\sum_{j=2}^K\sum_{i=j}^K(-1)^{i-j}\binom{i}{j}\frac{j^{N+1}-1}{j-1}=$$ $$=\sum_{j=2}^K\left[\frac{j^{N+1}-1}{2^{j+1}(j-1)}+\frac{1}{j-1}(-1)^{K-j} \left(j^{N+1}-1\right) \binom{K+1}{j} \, _2F_1(1,K+2;K+2-j;-1)\right]$$$$=\sum_{j=2}^K\frac{j^{N+1}-1}{j-1}\left[\frac{1}{2^{j+1}}+(-1)^{K-j} \binom{K+1}{j} \, _2F_1(1,K+2;K+2-j;-1)\right]$$
