Highest weight of conjugate representation I'm looking at the following problem

Let $\pi$ be an irreducible unitary representation of U(n) with highest weight $\lambda=(\lambda_1,\ldots,\lambda_n)$. Show that the highest weight of the conjugate representation $\bar{\pi}$ is $\lambda'=(-\lambda_n,\ldots,-\lambda_1)$.

A hint is given to show that $v'=\pi(w_0)w$ (for $w_0=\begin{pmatrix} & & 1 \\ & \dots & \\ 1 & &\end{pmatrix}$) is the highest weight vector of $\bar{\pi}$.
 A: The key point is that for a maximal torus $T\subset U(n)$ and $t\in T$, we have $\overline{t}=t^{-1}$. To see this, just look at the diagonal maximal torus $T\subset U(n)$, given by the set of matrices
$$T=\{diag(t_1,\ldots,t_n): t_i\overline{t_i}=1\}.$$ 
Now, let $V$ denote the space of the representation of highest weight $\lambda$ and let $v\in V$ be a weight vector of weight $\mu$. By definition, $\overline{\pi}(t):=\overline{\pi(t)}$, so we see that if $\pi(t)\cdot v= t^{\mu}v$, then $$ \overline{\pi}(t)\cdot v = \overline{\pi(t)}\cdot v = \overline{t^{\mu}}v=t^{-\mu}v.$$
Thus, $\mu$ is a weight of $\pi$ if and only if $-\mu$ is a weight of $\overline{\pi}$. 
Additionally, the symmetric group $S_n$ acts on the set of weights of $V$ as follows: for any permutation matrix $w\in S_n$, if the vector $v$ has weight $\mu=(\mu_1,\ldots,\mu_n)$, then $\pi(w)\cdot v$ has weight $w\mu=(\mu_{w(1)},\ldots,\mu_{w(n)})$. 
Now, let $w\in V$ be a highest weight vector, so that $\pi(t)\cdot w= t^{\lambda}v$. Then the element given in the problem $v'=\pi(w_0)w$ has weight $$w_0\lambda=(\lambda_n,\ldots,\lambda_1)$$
as a vector of the representation $(\pi,V)$.
Combining this with what we have seen about weights for the conjugate representation, we have that $v'$ has weight $-w_0\lambda = (-\lambda_n,\ldots, -\lambda_1)$ as a vector of the representation $(\overline{\pi},V)$. Now we just need to check that this weight is in fact a highest weight.
The only thing we need to note is that if $\mu$ and $\lambda$ are two weights and $\mu\leq \lambda$, then $w_0\lambda\leq w_0\mu$. I will leave this as a (straightforward) exercise in the definitions. 
Then, if $\mu$ is any weight of the representation $(\overline{\pi},V)$, we know that $w_0\mu$ is too, so that $-w_0\mu$ is a weight of the original representation $(\pi,V)$. Since $\lambda$ is the highest weight of $\pi$, we have
$$-w_0\mu\leq \lambda\Rightarrow w_0\lambda\leq -\mu\Rightarrow \mu\leq -w_0\lambda,$$
where I used the fact that $w_0^2=1$. This shows that $-w_0\lambda$ is a highest weight of $(\overline{\pi},V)$, and we are done.
