Proof By Induction With Integration Problem I am required to prove this formula by induction$$ \int x^k e^{\lambda x} = \frac{(-1)^{k+1}k!}{\lambda^{k+1}} + \sum_{i=0}^k \frac{(-1)^i k^\underline{i}}{\lambda^{i+1}}x^{k-i}e^{\lambda x}$$
where $k^{\underline i}$ is a falling factorial $k(k-1) \cdots (k-i+1)$ (assuming this to be equal to 1 when $i=0$) and the integral $\int f$ is defined as $\int_0^x f(\xi) d\xi$
My first problem results when trying for $k=0$ as an initial value, I get LHS: $\frac{e^{\lambda x}}{\lambda}$ and RHS: $\frac{e^{\lambda x}}{\lambda} - \frac{1}{\lambda}$ I assumed that as I have solved an integral the lambda on the RHS could be incorporated in the "+C" that the integral produced on the LHS but doesn't the $\int f$ definition mean that there is no +C?
Help would also be appreciated to any steps towards solving for $k=n+1$

According to your definition of the operator $\int$, the LHS for $k=0$ becomes $$\int e^{\lambda x}=\int_0^x e^{\lambda\xi}\; d\xi = \left[\frac{e^{\lambda\xi}}{\lambda}\right]_{\xi = 0}^{\xi = x}=\frac{e^{\lambda x}}{\lambda}-\frac1\lambda$$
And working out the right-hand side for $k=0$ will give you the same result.
