Help with $1 + a + a(a-1) + a(a-1) (a-2) +\cdots+a(a-1)\cdots(a-(n-1))$

Question

I want to rewrite the series $$1 + a + a(a-1) + a(a-1) (a-2) +\cdots+a(a-1)\cdots(a-(n-1))$$ as $(a^n-1)Y$ or $(a^{n-1}-1)Y$

Short-form: $$\{1+\sum_{i=1}^{n} \prod_{j=0}^{i-1}(a-j)\}$$ as $(a^n-1)Y$ or $(a^{n-1}-1)Y$

Answer 1

I don't think you can: $(a^n-1)Y$ is zero at $a=1$ but this is not true for $1 + a + a(a-1) + a(a-1) (a-2) + a(a-1)(a-2)(a-3)\cdots+a(a-1)\cdots (a-(n-1))$.

Answer 2

You can get a kind of recursive stuff: $$K(a):=1+a+a(a-1)+....+a(a-1)\cdot ...\cdot \left(a-(n-1)\right)=$$$$=1+a\left\{1+(a-1)\left[1+(a-2)\left(...(1+a-(n-1)\right)\right]\right\}$$This formula allows you to see, perhaps a little clearer than before, that$$K(1)=2\,\,,\,\,K(2)=5\,\,,\,K(3)=16\,\,,\,K(4)=66\,,...$$According to OEIS this is sequence $\,A000522\,$, which, pretty surprisingly for me, I must say, equals $$n!\sum_{k=0}^n\frac{1}{k!}$$with $\,K(0):=1\,$