Induction base case enquiry?

Question

$$\sum_{k=1}^n k(k+1)(k+2)...(k+p-1)= \frac{n(n+1)(n+2)...(n+p)} {p+1} $$

I am stuck on the base case part. I sub 1 into k and n but they don't yield the same value(R.H.S= 1*2*3...p /... L.H.S=1*2*3) could someone point out what I did wrong or is there an alternative way doing it. Thanks

Answer 1

For $n=1$, the only summand on the left is $1\times 2 \times \dots \times p = p!$ and on the right we have $$\frac{1\times 2\times\dots\times (p+1)}{p+1} = \frac{(p+1)!}{p+1} = p!$$