Let $n>3$ and $0<k<d-1$ be three natural numbers. Let the integer $a$ be the number of all non-negative integer solutions of the equation $ x_1+x_2+\dots+x_n=d $ which $k$ is not in the solutions. Also, set $b=d^{n-1}-(d-k)^{n-1}+n-1$. Prove that $a<b$. The other words for the problem is $$ d^{n-1}-(d-k)^{n-1}+n-1>c(n+d-1,d)-c(n+d-k-2,d-k).$$ Note that $c(m,r)=\dfrac{m!}{r!(m-r)!}$. It's well known that the number of all non-negative integer solutions of the equation $ x_1+x_2+\dots+x_n=d $ is equal to $c(n+d-1,d)$.
It may be useful to remark that the left side (the integer $b$) is equal to the cardinal of the following set. $$ \lbrace (a_1,\dots,a_{n-1})\in A \mid \exists j: a_j>d-k \rbrace \cup \lbrace 1,2,\dots,n-1 \rbrace $$ which $A=\lbrace 1,2,\dots,d \rbrace ^{n-1}$.