To show that $\sum_{x \in \lambda}(h(x)^2 - c(x)^2)=|\lambda|^2$, $h(x)$ is hook-length & $c(x)$ content of $x$, a block in the diagram of $\lambda$ I'm self-reading Macdonald's Symmetric Functions and Hall Polynomials, and he gives quite a few examples without any proof, and I'm unable to trace the line of thought' this is one of them. 
I have the following definitions and expressions at my disposal:
$x=(i,j)$ is a block in the Young Diagram of a partition $\lambda$ (with it's complement/transpose being $\lambda'$), and it's hook-length $h(x)$, and content $c(x)$ are defined as follows: 
$$h(x)=h(i,j)=\lambda_i+\lambda'_j - i - j +1 \\ c(x)=c(i,j)=j-i$$
I have been able to prove the following statements about these functions:
$$\sum_{x \in \lambda} h(x) = n(\lambda') + n(\lambda)+|\lambda| \\ \sum_{x \in \lambda} c(x) = n(\lambda') - n(\lambda)$$
where $n(\lambda)=\sum_{i \geq 1} (i-1)\lambda_i = \sum_{i \geq 1} \binom{\lambda'_i}{2}$
Now, I have to show that: $$\sum_{x \in \lambda} \big(h(x)^2 - c(x)^2\big) = |\lambda|^2$$
Example:
For example, let $\lambda = (1 \ 1) \implies \lambda'=(2)$
Then $ \ h(1,1) = 1 + 2 - 1 - 1 + 1 = 2 , \\ \ h(2,1) = 1 + 2 - 2 - 1 + 1 = 1$
And $ \ c(1,1) = 1-1 = 0, \\ \ c(2,1) = 1 - 2 = -1$
Hence, $$\sum_{x \in \lambda} \big(h(x)^2 - c(x)^2\big) = (2^2 - 0^2) + (1^2 - (-1)^2) = 2^2 = |\lambda|^2$$
 A: How about induction over $|\lambda|$? Clearly it holds for the unique partition of $1$, whose cell has hook $1$ and content $0$.
Suppose it holds for a partition $\lambda$ and consider the partition $\mu$ whose Ferrers diagram is equal to that of $\lambda$ but with the cell $(a, b)$ added. (For it to be a valid Ferrers diagram we therefore deduce that $\lambda_a = b-1$ and $\lambda'_b = a-1$).
Then $h_\mu(i,j) = h_\lambda(i,j)$ unless $i=a$ or $b=j$, in which case it is one greater. Therefore $$\begin{eqnarray*}\sum_{x \in \mu} \big(h_\mu(x)^2 - c(x)^2\big) & = & h_\mu(a, b)^2 - c(a, b)^2 + \sum_{x \in \lambda} \big(h_\mu(x)^2 - c(x)^2\big) \\
 & = & 1 - (a - b)^2 + \sum_{x \in \lambda} \big(h_\mu(x)^2 - c(x)^2\big) \\
 & = & 1 - (a - b)^2 + \sum_{x \in \lambda} \big((h_\lambda(x) + [i=a \vee b=j])^2 - c(x)^2\big) \\
 & = & 1 - (a - b)^2 + \sum_{x \in \lambda} \big(h_\lambda(x)^2 - c(x)^2\big) + \sum_{i=1}^{a-1} \big(2h_\lambda(i,b) + 1\big) + \sum_{j=1}^{b-1} \big(2h_\lambda(a,j) + 1\big) \\
 & = & |\lambda|^2 - (a - b)^2 + a + b - 1 + 2\left(\sum_{i=1}^{a-1} h_\lambda(i,b) + \sum_{j=1}^{b-1} h_\lambda(a,j)\right)
\end{eqnarray*}$$
Taking just the sums,
$$\begin{eqnarray*}
\sum_{i=1}^{a-1} h_\lambda(i,b) + \sum_{j=1}^{b-1} h_\lambda(a,j) & = & \sum_{i=1}^{a-1} \big(\lambda_i+\lambda'_b - i - b + 1\big) + \sum_{j=1}^{b-1} \big(\lambda_a+\lambda'_j - a - j + 1\big) \\
& = & \sum_{i=1}^{a-1} \big(\lambda_i - i + a - b\big) + \sum_{j=1}^{b-1} \big(\lambda'_j - j - a + b\big) \\
& = & (a-b)^2 + \sum_{i=1}^{a-1} \big(\lambda_i - i\big) + \sum_{j=1}^{b-1} \big(\lambda'_j - j\big) \\
& = & (a-b)^2 - \frac{(a-1)a + (b-1)b}{2} + \sum_{i=1}^{a-1} \lambda_i + \sum_{j=1}^{b-1} \lambda'_j \\
\end{eqnarray*}$$
Slight subtlety: I assert that $\sum_{i=1}^{a-1} \lambda_i + \sum_{j=1}^{b-1} \lambda'_j = |\lambda| + (a-1)(b-1)$ because by construction $(a,b)$ is just "outside" $\lambda$'s Ferrers diagram. Considering the situation geometrically we see that the sums count the entire Ferrers diagram, double-counting the cells which are both above and left of $(a,b)$. Switching back from geometric to algebraic we now get
$$\begin{eqnarray*}\sum_{x \in \mu} \big(h_\mu(x)^2 - c(x)^2\big) & = & |\lambda|^2 - (a - b)^2 + a + b - 1 + 2\left((a-b)^2 - \frac{(a-1)a + (b-1)b}{2} + |\lambda| + (a-1)(b-1)\right) \\
& = & |\lambda|^2 + 2|\lambda| + 1
\end{eqnarray*}$$
as desired.
Unfortunately this doesn't give much insight into the theorem. I'm sure there is a more elegant proof.
