Quick questions on Reduction formulae explained: H2 Further Mathematics

The standard derivation splits off one factor and integrates by parts. For $I_n = \int x^n\mathrm{e}^x\,dx$, take $u = x^n$ and $dv = \mathrm{e}^x\,dx$; integration by parts lowers the power of $x$ by one, producing $I_{n-1}$. For powers of sine, $\sin^n x = \sin^{n-1}x\cdot\sin x$, integrate by parts with $dv = \sin x\,dx$, then use $\cos^2 x = 1 - \sin^2 x$ to express the result back in terms of $I_n$ and $I_{n-2}$, and rearrange.

$\sin^n$ and $\cos^n$ reductions step by two, so an even $n$ ends at $I_0$ and an odd $n$ at $I_1$; using the wrong base case gives the wrong constant.

Always evaluate the $[uv]$ term; for $0$ to $\tfrac{\pi}{2}$ it usually vanishes, but check rather than assume.

$\sin^2 x = 1 - \cos^2 x$ introduces $-I_n$; mishandling the sign breaks the rearrangement.

State the reduction formula for $I_n = \int_0^{\pi/2}\sin^n x\,dx$. [1 mark]

What is the base case $I_0$ for $\int_0^{\pi/2}\sin^n x\,dx$? [1 mark]

Using $I_n = \mathrm{e} - n I_{n-1}$ with $I_0 = \mathrm{e} - 1$, find $I_1$. [1 mark]