This is how I would work the problem:
The product rule for a composite function that is the product of two functions is well-known and will be the basis for working this problem (and accepted without proof):
[MATH]\frac{d}{dx}\left(g_1(x)\cdot g_2(x) \right)=g_1'(x)\cdot g_2(x)+g_1(x)\cdot g_2'(x)[/MATH]
Using this rule, let's look at:
[MATH]\frac{d}{dx}\left(g_1(x)\cdot g_2(x)\cdot g_3(x) \right)[/MATH]
Now, let's associate two of the functions together, it doesn't matter which two, so let's use the first two:
[MATH]\frac{d}{dx}\left(\left(g_1(x)\cdot g_2(x) \right)\cdot g_3(x) \right)[/MATH]
Now, using the product rule above, we may state:
[MATH]\frac{d}{dx}\left(\left(g_1(x)\cdot g_2(x) \right)\cdot g_3(x) \right)=\frac{d}{dx}\left(g_1(x)\cdot g_2(x) \right)\cdot g_3(x)+\left(g_1(x)\cdot g_2(x) \right)\cdot g_3'(x)[/MATH]
Using the product rule again, we find:
[MATH]\frac{d}{dx}\left(\left(g_1(x)\cdot g_2(x) \right)\cdot g_3(x) \right)=\left(g_1'(x)\cdot g_2(x)+g_1(x)\cdot g_2'(x) \right)\cdot g_3(x)+\left(g_1(x)\cdot g_2(x) \right)\cdot g_3'(x)[/MATH]
And distributing, we find:
[MATH]\frac{d}{dx}\left(\left(g_1(x)\cdot g_2(x) \right)\cdot g_3(x) \right)=g_1'(x)\cdot g_2(x)\cdot g_3(x)+g_1(x)\cdot g_2'(x)\cdot g_3(x)+g_1(x)\cdot g_2(x)\cdot g_3'(x)[/MATH]
Now, this is enough to suggest the pattern (our induction hypothesis \(P_n\)):
[MATH]\frac{d}{dx}\left[\prod_{k=1}^n\left(g_k(x) \right) \right]=\sum_{k=1}^n\left[\prod_{j=1}^{k-1}\left(g_j(x) \right)\cdot\frac{d}{dx}\left(g_k(x) \right)\cdot\prod_{j=k+1}^n\left(g_j(x) \right) \right][/MATH]
Next, consider:
[MATH]\frac{d}{dx}\left[\prod_{k=1}^n\left(g_k(x) \right)\cdot g_{n+1}(x) \right][/MATH]
Using the product rule, and incorporating the new factor into the product. we may state:
[MATH]\frac{d}{dx}\left[\prod_{k=1}^{n+1}\left(g_k(x) \right) \right]=\frac{d}{dx}\left[\prod_{k=1}^n\left(g_k(x) \right) \right]\cdot g_{n+1}(x)+\prod_{k=1}^n\left(g_k(x) \right)\cdot g_{n+1}'(x)[/MATH]
Using our induction hypothesis, this becomes:
[MATH]\frac{d}{dx}\left[\prod_{k=1}^{n+1}\left(g_k(x) \right) \right]=\sum_{k=1}^n\left[\prod_{j=1}^{k-1}\left(g_j(x) \right)\cdot\frac{d}{dx}\left(g_k(x) \right)\cdot\prod_{j=k+1}^n\left(g_j(x) \right) \right]\cdot g_{n+1}(x)+\prod_{k=1}^n\left(g_k(x) \right)\cdot g_{n+1}'(x)[/MATH]
Now, incorporating the factor at the end of the first term on the right, we have:
[MATH]\frac{d}{dx}\left[\prod_{k=1}^{n+1}\left(g_k(x) \right) \right]=\sum_{k=1}^n\left[\prod_{j=1}^{k-1}\left(g_j(x) \right)\cdot\frac{d}{dx}\left(g_k(x) \right)\cdot\prod_{j=k+1}^{n+1}\left(g_j(x) \right) \right]+\prod_{k=1}^n\left(g_k(x) \right)\cdot g_{n+1}'(x)[/MATH]
And finally incorporating the second term on the right within the first summation term, we have:
[MATH]\frac{d}{dx}\left[\prod_{k=1}^{n+1}\left(g_k(x) \right) \right]=\sum_{k=1}^{n+1}\left[\prod_{j=1}^{k-1}\left(g_j(x) \right)\cdot\frac{d}{dx}\left(g_k(x) \right)\cdot\prod_{j=k+1}^{n+1}\left(g_j(x) \right) \right][/MATH]
We have derived \(P_{n+1}\) from \(P_n\), thereby completing the proof by induction.