Hello hp1357,
a) 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):
$$\frac{d}{dx}\left(g_1(x)\cdot g_2(x) \right)=g_1'(x)\cdot g_2(x)+g_1(x)\cdot g_2'(x)$$
Using this rule, let's look at:
$$\frac{d}{dx}\left(g_1(x)\cdot g_2(x)\cdot g_3(x) \right)$$
Now, let's associate two of the functions together, it doesn't matter which two, so let's use the first two:
$$\frac{d}{dx}\left(\left(g_1(x)\cdot g_2(x) \right)\cdot g_3(x) \right)$$
Now, using the product rule above, we may state:
$$\frac{d}{dx}\left(\left(g_1(x)\cdot g_2(x) \right)\cdot g_3(x) \right)=\frac{d}{dx}\left(g_1(x)\cdot g_2(x) \right)\cdot g_3(x)+\left(g_1(x)\cdot g_2(x) \right)\cdot g_3'(x)$$
Using the product rule again, we find:
$$\frac{d}{dx}\left(\left(g_1(x)\cdot g_2(x) \right)\cdot g_3(x) \right)=\left(g_1'(x)\cdot g_2(x)+g_1(x)\cdot g_2'(x) \right)\cdot g_3(x)+\left(g_1(x)\cdot g_2(x) \right)\cdot g_3'(x)$$
And distributing, we find:
$$\frac{d}{dx}\left(\left(g_1(x)\cdot g_2(x) \right)\cdot g_3(x) \right)=g_1'(x)\cdot g_2(x)\cdot g_3(x)+g_1(x)\cdot g_2'(x)\cdot g_3(x)+g_1(x)\cdot g_2(x)\cdot g_3'(x)$$
Now, this is enough to suggest the pattern (our induction hypothesis $P_n$):
$$\frac{d}{dx}\left[\prod_{k=1}^n\left(g_k(x) \right) \right]=\sum_{k=1}^n\left[\prod_{j=1}^{k-1}\left(g_j(x) \right)\cdot\frac{d}{dx}\left(g_k(x) \right)\cdot\prod_{j=k+1}^n\left(g_j(x) \right) \right]$$
Next, consider:
$$\frac{d}{dx}\left[\prod_{k=1}^n\left(g_k(x) \right)\cdot g_{n+1}(x) \right]$$
Using the product rule, and incorporating the new factor into the product. we may state:
$$\frac{d}{dx}\left[\prod_{k=1}^{n+1}\left(g_k(x) \right) \right]=\frac{d}{dx}\left[\prod_{k=1}^n\left(g_k(x) \right) \right]\cdot g_{n+1}(x)+\prod_{k=1}^n\left(g_k(x) \right)\cdot g_{n+1}'(x)$$
Using our induction hypothesis, this becomes:
$$\frac{d}{dx}\left[\prod_{k=1}^{n+1}\left(g_k(x) \right) \right]=\sum_{k=1}^n\left[\prod_{j=1}^{k-1}\left(g_j(x) \right)\cdot\frac{d}{dx}\left(g_k(x) \right)\cdot\prod_{j=k+1}^n\left(g_j(x) \right) \right]\cdot g_{n+1}(x)+\prod_{k=1}^n\left(g_k(x) \right)\cdot g_{n+1}'(x)$$
Now, incorporating the factor at the end of the first term on the right, we have:
$$\frac{d}{dx}\left[\prod_{k=1}^{n+1}\left(g_k(x) \right) \right]=\sum_{k=1}^n\left[\prod_{j=1}^{k-1}\left(g_j(x) \right)\cdot\frac{d}{dx}\left(g_k(x) \right)\cdot\prod_{j=k+1}^{n+1}\left(g_j(x) \right) \right]+\prod_{k=1}^n\left(g_k(x) \right)\cdot g_{n+1}'(x)$$
And finally incorporating the second term on the right within the first summation term, we have:
$$\frac{d}{dx}\left[\prod_{k=1}^{n+1}\left(g_k(x) \right) \right]=\sum_{k=1}^{n+1}\left[\prod_{j=1}^{k-1}\left(g_j(x) \right)\cdot\frac{d}{dx}\left(g_k(x) \right)\cdot\prod_{j=k+1}^{n+1}\left(g_j(x) \right) \right]$$
We have derived $P_{n+1}$ from $P_n$, thereby completing the proof by induction.
b) Now, if:
$$f(x)=\prod_{k=1}^n\left(g_k(x) \right)$$
and
$$g_k(x)=(1+kx)$$, we see that we have:
$$f'(x)=\sum_{k=1}^n\left[\prod_{j=1}^{k-1}\left(1+jx \right)\cdot k\cdot\prod_{j=k+1}^n\left(1+jx \right) \right]$$
Hence:
$$f'(0)=\sum_{k=1}^n(k)=\frac{n(n+1)}{2}$$
