Given is a partition $\lambda$, and the $\lambda$ also denote it's Young Diagram, and $\lambda'$ is the conjugate/transpose. Then the hook-length of $\lambda$ at $x = (i,j)$ is defined to be $$h(x) = h(i,j) = \lambda_i + \lambda'_j - i - j + 1$$We also have $$n(\lambda)=\sum_{i\geq 1} (i-1)\lambda_i=\sum_{i\geq 1}\binom{\lambda'_i}{2},\ \text{and}$$ $$|\lambda|=\sum_{i\geq 1}\lambda_i$$
I have to show that the sum of the hook-lengths of $\lambda$ is $$\sum_{x \in \lambda}h(x)=n(\lambda) + n(\lambda')+|\lambda|$$
My attempt: $$\sum_{x \in \lambda}h(x)=\sum_{i \geq 1}\sum_{j = 1}^{\lambda_i}h(i,j)=\sum_{i \geq 1}\sum_{j = 1}^{\lambda_i}(\lambda_i + \lambda'_j - i - j + 1)\\=\Big(\sum_{i \geq 1}\sum_{j = 1}^{\lambda_i}\lambda_j'\Big)+\sum_{i \geq 1}\Big(\lambda_i^{2}-i\lambda_i-\frac{\lambda_i(\lambda_i + 1)}{2}+\lambda_i\Big) \\ =\Big(\sum_{i \geq 1}\sum_{j = 1}^{\lambda_i}\lambda_j'\Big)+\sum_{i \geq 1}\bigg(\frac{\lambda_i(\lambda_i - 1)}{2}-(i-1)\lambda_i\bigg)\\=\Big(\sum_{i \geq 1}\sum_{j = 1}^{\lambda_i}\lambda_j'\Big)+n(\lambda') - n(\lambda)$$
I'm not really sure how to evaluate the first term of the final expression, and I'm getting a minus instead of a plus in the latter 2 terms; and hence I'm stuck. Can I get some kind of assistance?
$\lambda_j'$ is the number of parts in $\lambda$ that are greater than $j$. You can see that $$\sum_i\sum_{j=1}^{\lambda_i}\lambda'_j=\sum_{j \geq 1}(\lambda'_j)^2=\sum_{j \geq 1}\left(2\frac{\lambda'_j(\lambda'_j-1)}{2}+\lambda'_j\right)=2n(\lambda)+|\lambda|.$$
In the first step we have reversed the order of the sums by asking how many times each $\lambda_j'$ will appear. The answer is once for each $i$ such that $\lambda_i\ge j$ and this is precisely $\lambda_j'$ itself.